Grade 11 Mathematics: Quadratic Equation [BCB Exercise 6.2]

1. From the equation whose roots are
   1. 3,-2
      

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      Solution :1a
      Given roots roots are 3,-2, thus
      \( \text {sum of roots}=(3)+(-2)=1, \text {product of roots}=(3)(-2)=-6\)
      According to the question
      The new quadratic equation is
      \( x^2 - (\text {sum of roots}) x + (\text {product of roots}) =0\)
      or \( x^2- \left ( 1 \right) x + \left ( -6 \right )=0\)
      or \(x^2-x -6=0\)
      This is the required quadratic equation
      
      
   2. -5,4
      

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      Solution :1b
      Given roots roots are -5,4, thus
      \( \text {sum of roots}=(-5)+(4)=-1, \text {product of roots}=(-5)(4)=-20\)
      According to the question
      The new quadratic equation is
      \( x^2 - (\text {sum of roots}) x + (\text {product of roots}) =0\)
      or \( x^2- \left ( -1 \right) x + \left ( -20 \right )=0\)
      or \(x^2+x -20=0\)
      This is the required quadratic equation
      
      
   3. \(\sqrt{3},-\sqrt{3}\)
      

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      Solution :1c
      Given roots roots are \(\sqrt{3},-\sqrt{3}\), thus
      \( \text {sum of roots}=(\sqrt{3})+(-\sqrt{3})=0, \text {product of roots}=(\sqrt{3})(-\sqrt{3})=-3\)
      According to the question
      The new quadratic equation is
      \( x^2 - (\text {sum of roots}) x + (\text {product of roots}) =0\)
      or \( x^2- \left ( 0 \right) x + \left ( -3 \right )=0\)
      or \(x^2-3=0\)
      This is the required quadratic equation
      
      
   4. \(\frac{1}{2} (-1+\sqrt{5}),\frac{1}{2} (-1-\sqrt{5}) \)
      

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      Solution :1d
      Given roots roots are \(\frac{1}{2} (-1+\sqrt{5}),\frac{1}{2} (-1-\sqrt{5}) \), thus
      \( \text {sum of roots}=\left (\frac{1}{2} (-1+\sqrt{5}) +\frac{1}{2} (-1-\sqrt{5}) \right)=-1, \text {product of roots}=\left (\frac{1}{2} (-1+\sqrt{5}).\frac{1}{2} (-1-\sqrt{5}) \right)=-1\)
      According to the question
      The new quadratic equation is
      \( x^2 - (\text {sum of roots}) x + (\text {product of roots}) =0\)
      or \( x^2- \left ( -1 \right) x + \left ( -1 \right )=0\)
      or \(x^2+x-1=0\)
      This is the required quadratic equation
      
      
   5. -3+5i,-3-5i
      

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      Solution :1e
      Given roots roots are -3+5i,-3-5i, thus
      \( \text {sum of roots}=(-3+5i) +(-3+5i )=-6, \text {product of roots}=(-3+5i) (-3+5i )=34\)
      According to the question
      The new quadratic equation is
      \( x^2 - (\text {sum of roots}) x + (\text {product of roots}) =0\)
      or \( x^2- \left ( -6 \right) x + \left ( 34 \right )=0\)
      or \(x^2+6x+34=0\)
      This is the required quadratic equation
      
      
   6. a+ib,a-ib
      

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      Solution :1f
      Given roots roots are a+ib,a-ib, thus
      \( \text {sum of roots}=(a+ib) +(a-ib )=2a, \text {product of roots}=(a+ib) +(a-ib)=a^2+b^2\)
      According to the question
      The new quadratic equation is
      \( x^2 - (\text {sum of roots}) x + (\text {product of roots}) =0\)
      or \( x^2- \left ( 2a \right) x + \left ( a^2+b^2 \right )=0\)
      or \(x^2+2ax+a^2+b^2=0\)
      This is the required quadratic equation
      
      
2. 1. Find a quadratic equation whose roots are twice the roots of \(4x^2+8x-5=0\)
      

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      Solution :2a
      Given quadratic equation is
      \(4x^2+8x-5=0\)
      Let \( \alpha,\beta\) are the roots of \(4x^2+8x-5=0\), then
      \( \alpha+ \beta=-\frac{-8}{4}=-2, \alpha \beta=\frac{-5}{4}\)
      According to the question, \( 2\alpha,2\beta\) are the roots of new equation
      So, for the new quadratic equation
      \( \text {sum of roots}=(2\alpha)+ (2\beta)=2 (\alpha+ \beta)=-4\)
      \( \text {product of roots}=(2\alpha) (2\beta)=4 (\alpha \beta)=-5\)
      The new quadratic equation is
      \( x^2 - (\text {sum of roots}) x + (\text {product of roots}) =0\)
      or \( x^2- \left ( -4 \right) x + \left ( -5 \right )=0\)
      or \(x^2+4x-5=0\)
      This is the required quadratic equation
      
      
   2. Find a quadratic equation whose roots are reciprocals of the roots of \(3x^2-5x-2=0\)
      

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      Solution :2(b)
      Given quadratic equation is
      \(3x^2-5x-2=0\)
      Let \( \alpha,\beta\) are the roots of \(3x^2-5x-2=0\), then
      \( \alpha+ \beta=-\frac{5}{3}, \alpha \beta=\frac{-2}{3}\)
      According to the question, \( \frac{1}{\alpha},\frac{1}{\beta}\) are the roots of new equation
      So, for the new quadratic equation
      \( \text {sum of roots}=\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha + \beta}{\alpha \beta}=-\frac{5}{2}\)
      \( \text {product of roots}=\frac{1}{\alpha}\frac{1}{\beta}=\frac{1}{\alpha \beta}=-\frac{3}{2}\)
      The new quadratic equation is
      \( x^2 - (\text {sum of roots}) x + (\text {product of roots}) =0\)
      or \( x^2- \left ( - \frac{5}{2} \right) x + \left ( -\frac{3}{2} \right )=0\)
      or \(2x^2+5x-3=0\)
      This is the required quadratic equation
      
      
   3. Find a quadratic equation whose roots are greater by h than the roots of \(x^2-px+q=0\)
      

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      Solution :2(c)
      Given quadratic equation is
      \(x^2-px+q=0\)
      Let \( \alpha,\beta\) are the roots of \(x^2-px+q=0\), then
      \( \alpha+ \beta=p, \alpha \beta=q\)
      According to the question, \( \alpha +h,\beta +h\) are the roots of new equation
      So, for the new quadratic equation
      \( \text {sum of roots}=(\alpha+h)+(\beta+h)= 2h+ (\alpha+\beta)=2h+p\)
      \( \text {product of roots}=(\alpha+h)(\beta+h)= \alpha \beta+h(\alpha+\beta)+h^2=q+ph+h^2\)
      The new quadratic equation is
      \( x^2 - (\text {sum of roots}) x + (\text {product of roots}) =0\)
      or \( x^2- \left ( 2h+p \right) x + \left ( q+ph+h^2 \right )=0\)
      This is the required quadratic equation
      
      
   4. Find a quadratic equation whose roots are the squares of the roots of \(3x^2-5x-2=0\)
      

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      Solution :2(d)
      Given quadratic equation is
      \(3x^2-5x-2=0\)
      Let \( \alpha,\beta\) are the roots of \(3x^2-5x-2=0\), then
      \( \alpha+ \beta=\frac{5}{3}, \alpha \beta=-\frac{2}{3}\)
      According to the question, \( \alpha ^2,\beta ^2\) are the roots of new equation
      So, for the new quadratic equation
      \( \text {sum of roots}=(\alpha ^2)+(\beta ^2)= (\alpha+ \beta)^2-2 \alpha \beta=\left ( \frac{5}{3} \right )^2 -2 \frac{-2}{3}=\frac{37}{9}\)
      \( \text {product of roots}=(\alpha ^2)(\beta ^2)= (\alpha \beta)^2=\frac{4}{9}\)
      The new quadratic equation is
      \( x^2 - (\text {sum of roots}) x + (\text {product of roots}) =0\)
      or \( x^2- \left ( \frac{37}{9} \right) x + \left ( \frac{4}{9} \right )=0\)
      or \( 9x^2- 37 x + 4=0\)
      This is the required quadratic equation
      
      
3. Find a quadratic equation with rational coefficients one of whose roots is
   1. 4+3i
      

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      Solution :3a
      Given that
      one root is 4+3i
      We know that, immaginary roots always occur in conjugates, therefore
      other root is 4-3i
      
      
   2. \(\frac{1}{5+3i}\)
      

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      Solution :3b
      Given that
      one root is \(\frac{1}{5+3i}\)
      or one root is \(\frac{1}{5+3i} \frac{5-3i}{5-3i}=\frac{5-3i}{34}\)
      We know that, immaginary roots always occur in conjugates, therefore
      other root is \( \frac{5+3i}{34}\)
      
      
   3. \(2+\sqrt{3}\)
      

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      Solution :3c
      Given that
      one root is \(2+\sqrt{3}\)
      We know that, immaginary roots always occur in conjugates, therefore
      other root is \(2-\sqrt{3}\)
      
      
4. Find the value of k so that the equation
   1. \(2x^2+kx-15=0\) has one root 3
      

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      Solution :4a
      Given quadratic equation is
      \(2x^2+kx-15=0\)
      Comparing with \(ax^2+bx+c=0\), we get \(a=2,b=k,c=-15\)
      Also
      one root is 3, and suppose next root is \(\alpha\)
      Now
      \(3+ \alpha=-\frac{b}{a},3 \alpha =\frac{c}{a}\)
      or \(3+ \alpha=-\frac{k}{2},3 \alpha =\frac{-15}{2}\)
      or \(3+ \alpha=-\frac{k}{2},\alpha =\frac{-5}{2}\)
      or \(3-\frac{5}{2}=-\frac{k}{2}\)
      or \(\frac{1}{2}=-\frac{k}{2}\)
      or \(k=-1\)
      This completes the solution.
      
      
   2. \(3x^2+kx-2=0\) has roots whose sum is equal to 6
      

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      Solution :4b
      Given quadratic equation is
      \(3x^2+kx-2=0\)
      Comparing with \(ax^2+bx+c=0\), we get \(a=3,b=k,c=-2\)
      Also
      sum of roots is 6
      Now
      \(\alpha+\beta=6\)
      or \(-\frac{k}{3}=6\)
      or \(k=-18\)
      This completes the solution.
      
      
   3. \(2x^2+(4-k)x-17=0\) has roots equal but opposite in sign
      

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      Solution :4c
      Given quadratic equation is
      \(2x^2+(4-k)x-17=0\)
      Comparing with \(ax^2+bx+c=0\), we get \(a=2,b=4-k,c=-17\)
      Also
      roots are equal but opposite in sign
      Now
      \(\alpha=-\beta\)
      or \(\alpha +\beta =0\)
      or \(-\frac{b}{a}=0\)
      or \(b=0\)
      or \(4-k=0\)
      or \(k=4\)
      This completes the solution.
      
      
   4. \(3x^2+(5+k)x+8=0\) has roots numerically equal but opposite in sign
      

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      Solution :4d
      Given quadratic equation is
      \(3x^2+(5+k)x+8=0\)
      Comparing with \(ax^2+bx+c=0\), we get \(a=3,b=5+k,c=8\)
      Also
      roots are equal but opposite in sign
      Now
      \(b=0\)
      or \(5+k=0\)
      or \(k=-5\)
      This completes the solution.
      
      
   5. \(3x^2+7x+6-k=0\) has one root equal to zero
      

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      Solution :4e
      Given quadratic equation is
      \(3x^2+7x+6-k=0\)
      Comparing with \(ax^2+bx+c=0\), we get \(a=3,b=7,c=6-k\)
      Also
      one root is zero
      Now
      \(\alpha \beta =0\)
      or \(\frac{c}{a}=0\)
      or \(c=0\)
      or \(6-k=0\)
      or \(k=6\)
      This completes the solution.
      
      
   6. \(4x^2-17x+k=0\) has the reciprocal roots
      

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      Solution :4f
      Given quadratic equation is
      \(4x^2-17x+k=0\)
      Comparing with \(ax^2+bx+c=0\), we get \(a=4,b=-17,c=k\)
      Also
      roots are in reciprocal relation
      Now
      \(\alpha = \frac{1}{\beta }\)
      or \(\alpha \beta =1\)
      or \(\frac{c}{a}=1\)
      or \(\frac{k}{4}=1\)
      or \(k=4\)
      This completes the solution.
      
      
   7. \(4x^2+kx+5=0\) has roots whose difference is \(\frac{1}{4}\)
      

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      Solution :4g
      Given quadratic equation is
      \(4x^2+kx+5=0\)
      Comparing with \(ax^2+bx+c=0\), we get \(a=4,b=k,c=5\)
      Also
      difference of roots is 1/4
      or \(\alpha - \beta =\frac{1}{4}\)
      We know that
      \(\alpha + \beta =\frac{-k}{4},\alpha \beta =\frac{5}{4} \)
      Thus,
      \( (\alpha + \beta )^2= (\alpha- \beta )^2+ 4 \alpha \beta \)
      or \( \frac{k^2}{16}= \frac{1}{16}+5\)
      or \(k= \pm 9\)
      This completes the solution.
      
      
5. Show that -1 is a root of the equation \((a+b-2c)x^2+(2a-b-c)x+(c+a-2b)=0\). Find the other root.
   

   Solution 👉 Click Here

   Solution :5
   Given quadratic equation is
   \((a+b-2c)x^2+(2a-b-c)x+(c+a-2b)=0\)
   

   1. If (-1) is aa root, then it musst satisfy the equation
      \((a+b-2c)x^2+(2a-b-c)x+(c+a-2b)=0\)
      or \((a+b-2c)(-1)^2+(2a-b-c)(-1)+(c+a-2b)=0\)
      or \((a+b-2c)-(2a-b-c)+(c+a-2b)=0\)
      or \(0=0\)
      Therefore, we conclude that (-1) is a root.
   2. Given that (-1) is a root. Let us suppose that other root is \( \alpha\), then
      \( \alpha (-1)= \frac{c}{a}\)
      or \( \alpha (-1)= \frac{c+a-2b}{a+b-2c}\)
      or \( \alpha = - \frac{c+a-2b}{a+b-2c}\)
      this is the other root.

   This completes the solution.
6. Find the value of m for which the equation \((m+1)x^2+2(m+3)x+(2m+3)=0\) will have (a) reciprocal roots (b) one root zero.
   

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   Solution :6
   

   1. Given quadratic equation is
      \((m+1)x^2+2(m+3)x+(2m+3)=0\)
      Comparing with \(ax^2+bx+c=0\), we get \(a=m+1,b=2(m+3),c=2m+3\)
      Also
      roots are in reciprocal relation
      Now
      \(\alpha = \frac{1}{\beta }\)
      or \(\alpha \beta =1\)
      or \(\frac{c}{a}=1\)
      or \(\frac{2m+3}{m+1}=1\)
      or \(m=-2\)
      This completes the solution.
      
      
   2. Given quadratic equation is
      \((m+1)x^2+2(m+3)x+(2m+3)=0\)
      Comparing with \(ax^2+bx+c=0\), we get \(a=m+1,b=2(m+3),c=2m+3\)
      Also
      one root is zero
      Now
      \(\alpha \beta =0\)
      or \(\frac{c}{a}=0\)
      or \(\frac{2m+3}{m+1}=0\)
      or \(m=-3/2\)
      This completes the solution.
      
      
7. If the roots of the equation \(x^2+ax+c=0\) differ by 1, prove that \(a^2=4c+1\)
   

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   Solution :7
   Given quadratic equation is
   \(x^2+ax+c=0\)
   Let \(\alpha, \beta\) are the roots, then
   \( \alpha+ \beta =- \frac{b}{a} \) and \( \alpha \beta =\frac{c}{a} \)
   or \( \alpha+ \beta =-a \) and \( \alpha \beta =c \)
   According to the question
   \( \alpha- \beta =1\)
   or \( (\alpha- \beta)^2=1 \)
   or \( (\alpha+ \beta)^2-4 \alpha \beta =1 \)
   or \( a^2-4 c =1 \)
   or \( a^2=1 +4c\)
   This completes the proof.
   
   
8. If \( \alpha, \beta\) are the roots of the equation \(x^2-x-6=0\), find the equation whose roots are
   1. \(\alpha ^2 \beta ^{-1}\)and \(\beta ^2 \alpha ^{-1}\)
      

      Solution 👉 Click Here

      Solution :8a
      Given quadratic equation is
      \(x^2-x-6=0\)
      Let \( \alpha,\beta\) are the roots of \(x^2-x-6=0\), then
      \( \alpha+ \beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a}\)
      or \( \alpha+ \beta=1, \alpha \beta=-6\)
      According to the question, the new roots are
      \(\alpha ^2 \beta ^{-1}\) and \(\beta ^2 \alpha ^{-1}\)
      So,

      Sum of new roots	=\(\alpha ^2 \beta ^{-1}+ \beta ^2 \alpha ^{-1}\)
      	=\( \frac{\alpha ^2}{\beta}+ \frac{\beta ^2}{\alpha}\)
      	=\( \frac{\alpha ^3+\beta ^3}{\alpha \beta}\) =\( \frac{(\alpha +\beta) ^3-3\alpha \beta (\alpha +\beta) }{\alpha \beta}\) =\( \frac{(\alpha +\beta) ^3} {\alpha \beta} - \frac{3\alpha \beta (\alpha +\beta) }{\alpha \beta}\) =\( \frac{(\alpha +\beta) ^3} {\alpha \beta} - 3 (\alpha +\beta) \) =\( \frac{\left (1 \right ) ^3} {-6} - 3 \left (1 \right ) \) =\( -\frac{1}{6} -3 \) =\( -\frac{19}{6} \)

      Again,

      Product of new roots	=\(\alpha ^2 \beta ^{-1} . \beta ^2 \alpha ^{-1}\)
      	=\( \frac{\alpha ^2}{\beta}. \frac{\beta ^2}{\alpha}\)
      	=\( \alpha \beta\) =\( -6\)

      Now
      The new quadratic equation is
      \( x^2 - (\text {sum of roots}) x + (\text {product of roots}) =0\)
      or \( x^2- \left ( -\frac{19}{6} \right) x + \left (-6 \right)=0\)
      or \( 6x^2+19x -36=0\)
      This is the required quadratic equation
   2. \(\alpha + \frac{1}{\beta}\) and \(\beta + \frac{1}{\alpha}\)
      

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      Solution :8(b)
      Given quadratic equation is
      \(x^2-x-6=0\)
      Let \( \alpha,\beta\) are the roots of \(x^2-x-6=0\), then
      \( \alpha+ \beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a}\)
      or \( \alpha+ \beta=1, \alpha \beta=-6\)
      According to the question, the new roots are
      \(\alpha +\frac{1}{\beta}\) and \(\beta +\frac{1}{\alpha}\)
      So,

      Sum of new roots	=\(\alpha +\frac{1}{\beta}+\beta +\frac{1}{\alpha}\)
      	=\(\alpha +\beta +\frac{1}{\beta}+\frac{1}{\alpha} \)
      	=\( (\alpha +\beta) +\frac{\alpha +\beta}{\alpha \beta} \) =\( 1+\frac{1}{-6}\) =\(\frac{5}{6} \)

      Again,

      Product of new roots	=\( \left ( \alpha +\frac{1}{\beta} \right ). \left ( \beta +\frac{1}{\alpha}\right ) \)
      	=\(\frac{\alpha \beta+1}{\beta }\frac{\alpha \beta+1}{\alpha } \)
      	=\(\frac{(\alpha \beta)^2+2(\alpha \beta) +1}{\alpha \beta } \) =\(\frac{36-12+1}{-6}\) =\(-\frac{25}{6} \)

      Now
      The new quadratic equation is
      \( x^2 - (\text {sum of roots}) x + (\text {product of roots}) =0\)
      or \( x^2- \left ( \frac{5}{6} \right) x + \left (-\frac{25}{6} \right)=0\)
      or \( 6x^2-5x -25=0\)
      This is the required quadratic equation
9. If \( \alpha, \beta\) are the roots of the equation \(ax^2+bx+c=0\), find the equation whose roots are
   1. \(\alpha \beta ^{-1}\)and \(\beta \alpha ^{-1}\)
      

      Solution 👉 Click Here

      Solution :9a
      Given quadratic equation is
      \(ax^2+bx+c=0\)
      Let \( \alpha,\beta\) are the roots of \(ax^2+bx+c=0\), then
      \( \alpha+ \beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a}\)
      According to the question, the new roots are
      \(\alpha \beta ^{-1}\) and \(\beta \alpha ^{-1}\)
      So,

      Sum of new roots	=\(\alpha \beta ^{-1}+ \beta \alpha ^{-1}\)
      	=\( \frac{\alpha }{\beta}+ \frac{\beta }{\alpha}\)
      	=\( \frac{\alpha ^2+\beta ^2}{\alpha \beta}\) =\( \frac{(\alpha +\beta) ^2-2\alpha \beta }{\alpha \beta}\) =\( \frac{(\alpha +\beta) ^2}{\alpha \beta} -2 \frac {\alpha \beta }{\alpha \beta}\) =\( \frac{(\alpha +\beta) ^2}{\alpha \beta} -2 \) =\( \frac{b^2-2ac}{ac} \)

      Again,

      Product of new roots	=\(\alpha \beta ^{-1} . \beta \alpha ^{-1}\)
      	=\( \frac{\alpha }{\beta}. \frac{\beta }{\alpha}\)
      	=\( 1\)

      Now
      The new quadratic equation is
      \( x^2 - (\text {sum of roots}) x + (\text {product of roots}) =0\)
      or \( x^2- \left ( \frac{b^2-2ac}{ac} \right) x + \left (1 \right)=0\)
      or \( acx^2- (b^2-2ac)x + ac=0\)
      This is the required quadratic equation
   2. \(\alpha ^3 \) and \(\beta ^3\)
      

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      Solution :9b
      Given quadratic equation is
      \(ax^2+bx+c=0\)
      Let \( \alpha,\beta\) are the roots of \(ax^2+bx+c=0\), then
      \( \alpha+ \beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a}\)
      According to the question, the new roots are
      \(\alpha ^3 \) and \(\beta ^3\)
      So,

      Sum of new roots	=\(\alpha ^3+ \beta ^3\)
      	=\( (\alpha +\beta) ^3-3\alpha \beta (\alpha +\beta)\)
      	=\( \left (-\frac{b}{a} \right ) ^3- 3 \left (\frac{c}{a} \right ) \left (-\frac{b}{a} \right ) \) =\( -\frac{b^3}{a^3} + \frac{3bc}{a^2} \) =\( \frac{3abc-b^3}{a^3} \)

      Again,

      Product of new roots	=\(\alpha ^3 .\beta ^3\)
      	=\( \left ( \frac{c}{a} \right)^3\)
      	=\( \frac{c^3}{a^3}\)

      Now
      The new quadratic equation is
      \( x^2 - (\text {sum of roots}) x + (\text {product of roots}) =0\)
      or \( x^2- \left ( \frac{3abc-b^3}{a^3} \right) x + \left ( \frac{c^3}{a^3} \right)=0\)
      or \( a^3x^2+ (b^3-3abc)x + c^3=0\)
      This is the required quadratic equation
   3. \((\alpha-\beta)^2 \) and \((\alpha+\beta)^2 \)
      

      Solution 👉 Click Here

      Solution :9c
      Given quadratic equation is
      \(ax^2+bx+c=0\)
      Let \( \alpha,\beta\) are the roots of \(ax^2+bx+c=0\), then
      \( \alpha+ \beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a}\)
      According to the question, the new roots are
      \((\alpha-\beta)^2 \) and \((\alpha+\beta)^2 \)
      So,

      Sum of new roots	=\((\alpha+\beta)^2 +(\alpha-\beta)^2 \)
      	=\((\alpha+\beta)^2 +(\alpha+\beta)^2 -4 \alpha \beta \)
      	=\(2 \left ( \frac{b}{a}\right )^2 -4 \left ( \frac{c}{a}\right ) \) =\(\frac{2b^2-4ac}{a^2} \)

      Again,

      Product of new roots	=\((\alpha+\beta)^2 .(\alpha-\beta)^2 \)
      	=\((\alpha+\beta)^2 \left [(\alpha+\beta)^2 -4 \alpha \beta \right ]\)
      	=\((\alpha+\beta)^4 -4 (\alpha \beta )(\alpha+\beta)^2 \) =\(\frac{b^4}{a^4} -4 \frac{c}{a} \frac{b^2}{a^2} \) =\(\frac{b^4-4a^2b^2c}{a^4} \)

      Now
      The new quadratic equation is
      \( x^2 - (\text {sum of roots}) x + (\text {product of roots}) =0\)
      or \( x^2- \left ( \frac{2b^2-4ac}{a^2} \right) x + \left (\frac{b^4-4a^2b^2c}{a^4} \right)=0\)
      or \( a^4x^2- (2a^2b^2-4a^3c)x + (b^4-4a^2b^2c)=0\)
      This is the required quadratic equation
   4. the reciprocal of the roots of given equation
      

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      Solution :9d
      Given quadratic equation is
      \(ax^2+bx+c=0\)
      Let \( \alpha,\beta\) are the roots of \(ax^2+bx+c=0\), then
      \( \alpha+ \beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a}\)
      According to the question, the new roots are
      \( \frac{1}{\alpha} \) and \(\frac{1}{\beta} \)
      So,

      Sum of new roots	=\(\frac{1}{\alpha}+\frac{1}{\beta} \)
      	=\(\frac{\alpha+\beta}{\alpha \beta} \)
      	=\(- \frac{b}{c} \)

      Again,

      Product of new roots	=\(\frac{1}{\alpha} \frac{1}{\beta}\)
      	=\(\frac{1}{\alpha \beta}\)
      	=\(\frac{a}{c} \)

      Now
      The new quadratic equation is
      \( x^2 - (\text {sum of roots}) x + (\text {product of roots}) =0\)
      or \( x^2- \left ( -\frac{b}{c} \right) x + \left (\frac{a}{c} \right)=0\)
      or \( cx^2+bx +a=0\)
      This is the required quadratic equation
10. 1. If the roots of the equation \(ax^2+bx+c=0\) be in the ratio of 3:4, prove that \(12b^2=49ac\)
       

       Solution 👉 Click Here

       Solution :10a
       Given quadratic equation is
       \(ax^2+bx+c=0\)
       Assume that \( 3k,4k\) are the roots of \(ax^2+bx+c=0\), then
       \( 3k+ 4k=-\frac{b}{a}, 3k.4k=\frac{c}{a}\)
       or \( 7k=-\frac{b}{a}, 12k^2=\frac{c}{a}\)
       or \( k=-\frac{b}{7a}, 12k^2=\frac{c}{a}\)
       or \( 12 \left ( -\frac{b}{7a} \right )^2=\frac{c}{a}\)
       or \( 12 \left ( \frac{b^2}{49a^2} \right )=\frac{c}{a}\)
       or \( 12 b^2=49ac\)
       This completes the proof.
       
       
    2. If one root of the equation \(ax^2+bx+c=0\) be four times the other root, show that \(4b^2=25ac\)
       

       Solution 👉 Click Here

       Solution :10b
       Given quadratic equation is
       \(ax^2+bx+c=0\)
       Assume that \( k,4k\) are the roots of \(ax^2+bx+c=0\), then
       \( k+ 4k=-\frac{b}{a}, k.4k=\frac{c}{a}\)
       or \( 5k=-\frac{b}{a}, 4k^2=\frac{c}{a}\)
       or \( k=-\frac{b}{5a}, 4k^2=\frac{c}{a}\)
       or \( 4 \left ( -\frac{b}{5a} \right )^2=\frac{c}{a}\)
       or \( 4 \left ( \frac{b^2}{25a^2} \right )=\frac{c}{a}\)
       or \( 4 b^2=25ac\)
       This completes the proof.
       
       
    3. For what values of m, the equation \(x^2-mx+m+1=0\) may have its root in the ratio 2:3
       

       Solution 👉 Click Here

       Solution :10c
       Given quadratic equation is
       \(x^2-mxx+m+1=0\)
       Assume that \( 2k,3k\) are the roots of \(x^2-mxx+m+1=0\), then
       \( 2k+ 3k=m, 2k.3k=m+1\)
       Now, using the roots, we get
       \(6k^2=m+1\)
       or\(6 \left (\frac{m}{5} \right )^2 =m+1\)
       or\(\frac{6m^2}{25} =m+1\)
       or\(6m^2-25m-25=0\)
       or\(m=5,-\frac{5}{6}\)
       This completes the proof
       
       
11. 1. If \( \alpha, \beta\) are the roots of the equation \(px^2+qx+q=0\), prove that \(\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}+\sqrt{\frac{q}{p}}=0\)
       

       Solution 👉 Click Here

       Solution :11a
       

       Method 1

       Given quadratic equation is
       \(px^2+qx+q=0\)
       Given that \( \alpha, \beta\) are the roots of \(px^2+qx+q=0\), then
       \( \alpha+ \beta=-\frac{q}{p}, \alpha \beta=\frac{q}{p}\)
       Now, using the roots, we get
       \(\frac{\alpha+ \beta}{\sqrt{\alpha \beta}}=\frac{-\frac{q}{p}}{\frac{q}{p}}\)
       or\(\frac{\alpha}{\sqrt{\alpha \beta}}+\frac{ \beta}{\sqrt{\alpha \beta}}=- \sqrt{\frac{q}{p}}\)
       or\( \sqrt{ \frac{\alpha}{\beta}}+ \sqrt{ \frac{ \beta}{\alpha }}+ \sqrt{\frac{q}{p}}=0\)
       This completes the proof
       
       

       Method 2

       Given quadratic equation is
       \(px^2+qx+q=0\)
       Given that \( \alpha, \beta\) are the roots of \(px^2+qx+q=0\), then
       \( \alpha+ \beta=-\frac{q}{p}, \alpha \beta=\frac{q}{p}\)
       Now, taking LHS, we have
       \( \sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}+\sqrt{\frac{q}{p}}\)
       or\( \frac{\sqrt{\alpha}}{\sqrt{\beta}}+\frac{\sqrt{\beta}}{\sqrt{\alpha}}+\sqrt{\frac{q}{p}}\)
       or\( \frac{\alpha+\beta}{\sqrt{\alpha \beta}}+\sqrt{\frac{q}{p}}\)
       or\( \frac{-\frac{q}{p}}{\sqrt{\frac{q}{p}}}+\sqrt{\frac{q}{p}}\)
       or\( -\sqrt{\frac{q}{p}}+\sqrt{\frac{q}{p}}\)
       or0
       Which is RHS, therefore, we have
       \( \sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}+\sqrt{\frac{q}{p}}=0\)
       
       
    2. If roots of the equation \(lx^2+nx+n=0\) be in the ratio of p:q, prove that \(\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}=0\)
       

       Solution 👉 Click Here

       Solution :11b
       

       Method 0

       Given quadratic equation is
       \(lx^2+nx+n=0\)
       Given that \( \alpha, \beta\) are the roots of \(lx^2+nx+n=0\), then
       \( \alpha+ \beta=-\frac{n}{l}, \alpha \beta=\frac{n}{l}, \frac{\alpha}{\beta}=\frac{p}{q}\)
       Now, using the roots, we get
       \(\frac{\alpha+ \beta}{\sqrt{\alpha \beta}}=\frac{-\frac{n}{l}}{ \sqrt{\frac{n}{l}}}\)
       or\(\frac{\alpha}{\sqrt{\alpha \beta}}+\frac{ \beta}{\sqrt{\alpha \beta}}=- \sqrt{\frac{n}{l}}\)
       or\( \sqrt{ \frac{\alpha}{\beta}}+ \sqrt{ \frac{ \beta}{\alpha }}+ \sqrt{\frac{n}{l}}=0\)
       or\( \sqrt{ \frac{p}{q}}+ \sqrt{ \frac{ q}{p}}+ \sqrt{\frac{n}{l}}=0\)
       This completes the proof
       
       

       Method 1

       
       Given quadratic equation is
       \(lx^2+nx+n=0\)
       Let us assume that \( p \alpha,q \alpha \) are the roots of \(lx^2+nx+n=0\), then the roots are in the ration of p:q, now according to the question
       \(p \alpha q \alpha =\frac{n}{l}\)
       or \(pq \alpha ^2 =\frac{n}{pql}\)
       Since \(p \alpha\) is the one root of the equation \(lx^2+nx+n=0\), we get
       \(l (p \alpha)^2+n(p \alpha)+n=0\)
       or \(lp^2 (\alpha ^2) +np(\alpha)+n=0\)
       or \(lp^2 \frac{n}{pql} +np \sqrt{\frac{n}{pql}} +n=0\)
       or \(\frac{p}{q} +\sqrt{\frac{np}{ql}} +1=0\)
       Dividing both side by \(\sqrt{\frac{p}{q}} \), we get
       \(\sqrt{\frac{p}{q}} +\sqrt{\frac{n}{l}} +\sqrt{\frac{q}{p}}=0\)
       or \(\sqrt{\frac{p}{q}} +\sqrt{\frac{q}{p}} +\sqrt{\frac{n}{l}} =0\)
       
       
       

       Method 2

       Given quadratic equation is
       \(lx^2+nx+n=0\)
       Let \( \alpha, \beta\) are the roots of \(lx^2+nx+n=0\), then
       \( \alpha+ \beta=-\frac{n}{l}, \alpha \beta=\frac{n}{l},\frac{\alpha}{ \beta}=\frac{p}{q}\)
       Now, taking LHS, we have
       \( \sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}\)
       or\( \sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}+\sqrt{\frac{n}{l}}\)
       or\( \frac{\sqrt{\alpha}}{\sqrt{\beta}}+\frac{\sqrt{\beta}}{\sqrt{\alpha}}+\sqrt{\frac{n}{l}}\)
       or\( \frac{\alpha+\beta}{\sqrt{\alpha \beta}}+\sqrt{\frac{n}{l}}\)
       or\( \frac{-\frac{n}{l}}{\sqrt{\frac{n}{l}}}+\sqrt{\frac{n}{l}}\)
       or\( -\sqrt{\frac{n}{l}}+\sqrt{\frac{n}{l}}\)
       or0
       Which is RHS, therefore, we have
       \( \sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}=0\)
       
       
12. If one root of the equation \(ax^2+bx+c=0\) be square of the other root, prove that \(b^3+a^2c+ac^2=3abc\)
    

    Solution 👉 Click Here

    Solution :12
    Given quadratic equation is
    \(ax^2+bx+c=0\)
    Let us assume that \( \alpha,\beta\) are the roots of \(ax^2+bx+c=0\), then
    According to question
    \(\alpha=\beta ^2, \alpha+\beta =-\frac{b}{a}, \alpha\beta =\frac{c}{a}\)
    or \(\beta ^2+\beta =-\frac{b}{a}, \beta ^2\beta =\frac{c}{a}\)
    or \(\beta (1+ \beta) =-\frac{b}{a}, \beta ^3 =\frac{c}{a}\)
    Taking the cube of first part, we get
    \(\beta ^3 (1+ \beta)^3 =-\frac{b^3}{a^3}\)
    or \(\frac{c}{a} (1+ 3 \beta+3 \beta ^2+\beta ^3) =-\frac{b^3}{a^3}\)
    or \(\frac{c}{a} (1+ 3 \beta (1+\beta) +\beta ^3) =-\frac{b^3}{a^3}\)
    or \(\frac{c}{a} \left ( 1+ 3 \left ( -\frac{b}{a} \right ) +\frac{c}{a} \right ) =-\frac{b^3}{a^3}\)
    or \(\frac{c}{a} \left ( \frac{a-3b+c}{a} \right ) =-\frac{b^3}{a^3}\)
    or \(\frac{ac-3bc+c^2}{a^2} =-\frac{b^3}{a^3}\)
    or \(a^2c-3abc+ac^2 =-b^3\)
    or \(b^3+a^2c+ac^2 =3abc\)