Mathematics · Quantitative Aptitude

Polynomial and Quadratic Equations

239 Questions

Polynomial and quadratic equations involve finding the roots of expressions with varying degrees. The focus is on the relationship between coefficients and roots, symmetric functions, and solving higher degree polynomials. This topic is a core component of algebra in competitive testing.

Roots of quadratic equationsSymmetric root functionsCubic polynomialsRoot approximation methodsSum and product of roots

Polynomial and Quadratic Equations Questions

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\dfrac{x^2 - bx}{ax - c} = \dfrac{m - 1}{m + 1}$ has roots which are numerically equal but of opposite sings, the value of m must be:

  1. $\dfrac{a-b}{a + b}$
  2. $\dfrac{a + b}{a - b}$

  3. c
  4. $\dfrac{1}{c}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\Longrightarrow \cfrac { { x }^{ 2 }-bx }{ ax-c } =\cfrac { m-1 }{ m+1 } \ \Longrightarrow (m+1){ x }^{ 2 }-b(m+1)x=(m-1)ax-c(m-1)\ \Longrightarrow (m+1){ x }^{ 2 }-[b(m+1)+(m-1)a]x+c(m-1)=0$

Roots are numerically equal but of opposite sign.
$\therefore$ Sum of roots = 0
$\Longrightarrow (b+a)m+(b-a)=0$ 
$\therefore m=\cfrac { a-b }{ a+b } $

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\alpha$ and $\beta$ are the roots of the equation $ax^{2} \, + \, bx \, + \, c \, = \, 0$. The equation whose roots are as given below.
$\dfrac{\alpha }{\beta } \, ,\dfrac{\beta }{\alpha}$ is $acx^2 \, - \, (b^2 \, - \, 2ac) \, x \, + \, ac \, = \, 0$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\Rightarrow$  $\alpha$ and $\beta$ are roots of the equation $ax^2+bx+c=0$

$\Rightarrow$  $\alpha\beta=\dfrac{c}{a}$           ------ ( 1 )
$\Rightarrow$  $\alpha+\beta=\dfrac{-b}{a}$         ----- ( 2 )
$\Rightarrow$  $(\alpha+\beta)^2=\alpha^2+\beta^2+2\alpha\beta$
Using ( 1 ) and ( 2 ),
$\Rightarrow$  $\left(\dfrac{-b}{a}\right)^2=\alpha^2+\beta^2+2\times \dfrac{c}{a}$

$\Rightarrow$  $\dfrac{b^2}{a^2}=\alpha^2+\beta^2+\dfrac{2c}{a}$

$\therefore$   $\alpha^2+\beta^2=\dfrac{b^2}{a^2}-\dfrac{2c}{a}$

$\therefore$  $\alpha^2+\beta^2=\dfrac{b^2-2ac}{a^2}$          ------ ( 3 )
Now,
$\Rightarrow$  $\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}=\dfrac{\alpha^2+\beta^2}{\alpha\beta}$

                     $=\dfrac{\dfrac{b^2-2ac}{a^2}}{\dfrac{c}{a}}$           [ Using ( 1 ) and ( 3 ) ]

$\Rightarrow$  $\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}=\dfrac{b^2-2ac}{ac}$           ----- ( 4 )

$\Rightarrow$  $\dfrac{\alpha}{\beta}.\dfrac{\beta}{\alpha}=1$   ---- ( 5 )
New equation,
$\Rightarrow$  $x^2-\left(\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}\right)x+\left(\dfrac{\alpha}{\beta}.\dfrac{\beta}{\alpha}\right)=0$
By Using ( 4 ) and ( 5 ),
$\Rightarrow$  $x^2-\left(\dfrac{b^2-4ac}{ac}\right)x+1=0$

$\Rightarrow$  $acx^2-(b^2-4ac)x+1=0$
$\therefore$  We can see equation given in question is correct.

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If the roots of a quadratic equation are reciprocals of the roots of $ax^2 + bx + c = 0$, then what will be the coefficient of $c$?

  1. $x$
  2. $x^2$
  3. $-x$
  4. $x^3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know to find the equation with reciprocal roots, we simply put $\dfrac {1}{x}$ in the place of $x$.
so, $a(\dfrac {1}{x})^2 + b(\dfrac {1}{x}) + c = 0$
On taking LCM we get the required equation
$a + bx + cx^2 = 0$
So the coefficient of $c$ is $x^2$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

Find the Quadratic Equation whose roots are Reciprocal of $ax^2 + bx + c = 0$.

  1. $ax^2 + bx + c = 0$
  2. $a(\dfrac {1}{x})^2 + b(\dfrac {1}{x}) + c = 0$
  3. $3ax^2 + 2bx + c = 0$

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that for reciprocal roots, we only need to replace $x$ by $\dfrac {1}{x}$, in the given equation.
So the above equation becomes:
$a(\dfrac {1}{x})^2 + b(\dfrac {1}{x}) + c = 0$, which is the required answer.

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If A.M. of the roots of a quadratic equation is $8/5$ and A.M. of their reciprocals is $8/7$, then the equation is?

  1. $7x^2-16x+8=0$
  2. $3x^2-12x+7=0$
  3. $5x^2-16x+7=0$
  4. $7x^2-16x+5=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\Rightarrow$  Let $\alpha$ and $\beta$ are the roots of the equation.

According to the question,
$\Rightarrow$ $\dfrac{\alpha+\beta}{2}=\dfrac{8}{5}$

$\therefore$  $\alpha+\beta=\dfrac{16}{5}$                ----- ( 1 )

According to the equation,
$\Rightarrow$  $\dfrac{\dfrac{1}{\alpha}+\dfrac{1}{\beta}}{2}=\dfrac{8}{7}$
$\Rightarrow$  $\dfrac{\alpha+\beta}{2\alpha\beta}=\dfrac{8}{7}$
$\Rightarrow$  $7(\alpha+\beta)=16\alpha\beta$
$\Rightarrow$  $\dfrac{7\times \dfrac{16}{5}}{16}=\alpha\beta$
$\therefore$    $\alpha\beta=\dfrac{7}{5}$          - ------ ( 2 )
Now, new eqution,
$\Rightarrow$  $x^2-(\alpha+\beta)x+(\alpha.\beta)=0$
From ( 1 ) and ( 2 ),
$\Rightarrow$  $x^2-\dfrac{16}{5}x+\dfrac{7}{5}=0$
$\Rightarrow$  $5x^2-16x+7=0$   

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\alpha, \beta$ are the root of a quadratic equation $x^2 - 3x+5=0$, then the equation whose roots are $(\alpha^2 - 3 \alpha +7)$ and $(\beta^2 -3\beta +7)$ is

  1. $x^2 +4x+1=0$
  2. $x^2 -4x+4=0$
  3. $x^2 -4x-1=0$
  4. $x^2 +2x+3=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since $\alpha, \beta$ are the root of equation $x^2-3x+5=0$
So, $\alpha^2-3\alpha +5=0$
$\beta^2 -3\beta +5=0$
$\therefore \alpha^2 -3\alpha =-5$
$\beta^2 - 3 \beta =-5$
Putting in $(\alpha^2 - 3 \alpha +7) $  &  $(\beta^2 - 3\beta +7)$    ....... (1)
$-5 +7, -5 +7$
$\therefore$ 2 and 2 are the roots
$\therefore$ The required equation is $x^2 - 4x+4=0$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If the roots of ${a _1}{x^2}\, + \,{b _1}x\, + \,{c _1}\, = \,0$ are ${\alpha _1},\,{\beta _1},\,$ and those of ${a _2}{x^2}\, + \,{b _2}x\, + {c _2}\, = \,0$ are ${\alpha _2}\,,{\beta _2}$ such that ${\alpha _1}\,{\alpha _2} = \,{\beta _1}\,{\beta _2}\, = \,1$, then

  1. $\dfrac{{{a _1}}}{{{a _2}}} = \,\dfrac{{{b _1}}}{{{b _2}}}\, = \,\dfrac{{{c _1}}}{{{c _2}}}$
  2. $\dfrac{{{a _1}}}{{{c _2}}} = \,\dfrac{{{b _1}}}{{{b _2}}}\, = \,\dfrac{{{c _1}}}{{{a _2}}}$
  3. ${a _1}\,{a _2}\, = \,{b _1}\,{b _2}\, = \,{c _1}\,{c _2}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$a _1x^2+b _1x+c _1=0$

$\alpha _1+\beta _1=\cfrac{-b _1}{a _1}$
$\alpha _1\beta _1=\cfrac{c _1}{a _1}$

$a _2x^2+b _2x+c _2=0$
$\alpha _2+\beta _2=\cfrac{-b _2}{a _2}$
$\alpha _2\beta _2=\cfrac{c _2}{a _2}$

$\alpha _1\alpha _2=\beta _1\beta _2=1$
$\therefore \alpha _1\beta _1\alpha _2\beta _2=(\alpha _1\alpha _1)\cdot (\beta _1\beta _1)=1\cdot 1=1$
$\therefore \alpha _1\beta _1\alpha _2\beta _2=(\alpha _1\alpha _1)\cdot (\beta _1\beta _1)=\cfrac{c _1}{a _1}\cdot \cfrac{c _2}{a _2}=1$
$\implies \cfrac{c _1}{a _2}=\cfrac{a _1}{c _2}$
Now, $\alpha _1+\beta _1=\cfrac{-b _1}{a _1}$
$\implies \cfrac{1}{\alpha _2}+\cfrac{1}{\beta _2}=\cfrac{-b _1}{a _1}$
$\implies \cfrac{\alpha _2+\beta _2}{\alpha _2\beta _2}=\cfrac{-b _1}{a _1}$
$\implies \cfrac{-b _2/a _2}{c _2/a _2}=\cfrac{-b _1}{a _1}$
$\implies \cfrac{b _2}{c _2}=\cfrac{b _1}{a _1}$
$\implies \cfrac{a _1}{c _2}=\cfrac{b _1}{b _2}$
$\therefore \cfrac{a _1}{a _2}=\cfrac{b _1}{b _2}=\cfrac{c _1}{c _2}$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $alpha, beta$ are roots of $Ax^2 + Bx + C = 0$ and $\alpha^2, \beta^2$ are roots of $x^2 + px + q = 0$, the $p$ is equal to

  1. $\dfrac{B^2 - 2AC}{A^2}$
  2. $\dfrac{2AC - B^2}{A^2}$
  3. $\dfrac{B^2 - 4AC}{A^2}$
  4. $\dfrac{4AC - B^2}{A^2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$Ax^2+Bx+C=0$
$\alpha +\beta =\dfrac{-B}{A}$
$\alpha\beta =\dfrac{C}{A}$
Now it roots are $\alpha^2$ & $\beta^2$
Then equation will be
$(x-\alpha^2)(x-\beta^2)=0$
$x^2-x\beta^2-x\alpha^2+\alpha^2\beta^2=0$
$x^2-)(\alpha^2+\beta^2)x+\alpha^2\beta^2=0$
$x^2+[-(\alpha^2+\beta^2)]x+\alpha^2\beta^2=0$
$\alpha +\beta =\dfrac{-B}{-A}$
$\Rightarrow \alpha^2+\beta^2+2\alpha\beta =\dfrac{B^2}{A^2}$
$\alpha^2+\beta^2=\dfrac{B^2}{A^2}-\dfrac{2C}{A}$
$-(\alpha^2+\beta^2)=\dfrac{2C}{A}-\dfrac{B^2}{A^2}=\dfrac{2AC-B^2}{A^2}$
$p=\dfrac{2AC-B^2}{A^2}$.
Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\alpha+\beta$$=-2$ and ${\alpha}^{3}+{\beta}^{3}$$=-56$ then the quadratic equation whose roots are $\alpha,\beta$ is 

  1. ${ x }^{ 2 }+2x-16$$=0$
  2. ${x}^{2}+2x-15$$=0$
  3. ${x}^2+2x-12$$=0$
  4. ${x}^{2}+2x-8$$=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\Rightarrow$  $\alpha+\beta=-2$             ------ ( 1 )


$\Rightarrow$  $\alpha^3+\beta^3=-56$


$\Rightarrow$  $(\alpha+\beta)^3=\alpha^3+\beta^3+3\alpha^2\beta+3\alpha\beta^2$

$\Rightarrow$  $(\alpha+\beta)^3=\alpha^3+\beta^3+3\alpha\beta(\alpha+\beta)$

$\Rightarrow$  $(-2)^3=-56+3\alpha\beta(-2)$            [ Using ( 1 ) and ( 2 ) ]

$\Rightarrow$  $-8+56=-6\alpha\beta$

$\Rightarrow$  $48=-6\alpha\beta$

$\Rightarrow$  $\alpha\beta=-8$                      ----- ( 2 )

The required quadratic equation,

$x^2-(\alpha+\beta)x+(\alpha\beta)=0$

Using ( 1 ) and ( 3 ) we get,
$\Rightarrow$  $x^2+2x-8=0$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\alpha \neq \beta$ but $\alpha^2 = 5 \alpha -3$ and $\beta^2 = 5\beta -3$, then the equation whose roots are $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$is

  1. $3x^2 - 25x+3=0$
  2. $x^2 +5x-3=0$
  3. $x^2 -5x+3=0$
  4. $3x^2 - 19x+3=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
${\alpha}^{2}=5\alpha-3$

${\alpha}^{2}-5\alpha+3=0$

$\alpha=\dfrac{5\pm\sqrt{{5}^{2}-4\times 1\times 3}}{2}$

$\alpha=\dfrac{5\pm\sqrt{25-12}}{2}$

$\alpha=\dfrac{5\pm\sqrt{13}}{2}$

${\beta}^{2}=5\beta-3$

${\beta}^{2}-5\beta+3=0$

$\beta=\dfrac{5\pm\sqrt{{5}^{2}-4\times 1\times 3}}{2}$

$\beta=\dfrac{5\pm\sqrt{25-12}}{2}$

$\beta=\dfrac{5\pm\sqrt{13}}{2}$

Given:$\alpha\neq\,\beta$

Let $\alpha=\dfrac{5+\sqrt{13}}{2}$ and $\beta=\dfrac{5-\sqrt{13}}{2}$

$\Rightarrow\,\dfrac{\alpha}{\beta}=\dfrac{\dfrac{5+\sqrt{13}}{2}}{\dfrac{5-\sqrt{13}}{2}}$

$=\dfrac{5+\sqrt{13}}{5-\sqrt{13}}$

$=\dfrac{5+\sqrt{13}}{5-\sqrt{13}}\times \dfrac{5+\sqrt{13}}{5+\sqrt{13}}$

$=\dfrac{25+13+10\sqrt{13}}{25-13}$

$=\dfrac{38+10\sqrt{13}}{12}$

$=\dfrac{19+5\sqrt{13}}{6}$

$\Rightarrow\,\dfrac{\beta}{\alpha}=\dfrac{\dfrac{5-\sqrt{13}}{2}}{\dfrac{5+\sqrt{13}}{2}}$

$=\dfrac{5-\sqrt{13}}{5+\sqrt{13}}$

$=\dfrac{5-\sqrt{13}}{5+\sqrt{13}}\times \dfrac{5-\sqrt{13}}{5-\sqrt{13}}$

$=\dfrac{25+13-10\sqrt{13}}{25-13}$

$=\dfrac{19-10\sqrt{13}}{12}$

$=\dfrac{19-5\sqrt{13}}{6}$

Sum of the zeroes$=\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}$

$=\dfrac{19+5\sqrt{13}}{6}+\dfrac{19-5\sqrt{13}}{6}$

$=\dfrac{19+5\sqrt{13}+19-5\sqrt{13}}{6}$

$=\dfrac{2\times 19}{6}=\dfrac{19}{3}$

Product of the zeroes$=\dfrac{\alpha}{\beta}\times\dfrac{\beta}{\alpha}$

$=\dfrac{19+5\sqrt{13}}{6}\times\dfrac{19-5\sqrt{13}}{6}$

$=\dfrac{361-25\times 13}{36}=\dfrac{361-325}{36}=\dfrac{36}{36}=1$

Now,we know the foumula for finding quadratic equations,
${x}^{2}-\left(sum\,of \,the \,zeroes\right)x+product\,of\,the \,zeroes=0$

${x}^{2}-\dfrac{19}{3}x+1=0$

Hence the equation is $3{x}^{2}-19x+3=0$
Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If the difference of the roots of the quadratic equation is 3 and difference between their cubes is 189, then the quadratic equation is x2±9x+18=0x2±9x+18=0
State true or false.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the roots of the equation be a and b
then, $a^3 - b^3 = 189$
and $a - b = 3$
cubing both sides:
$(a-b)^3 = 27 $
$a^3 - b^3 - 3ab (a-b) = 27$
$189 -3ab(3) = 27 $
$162 = 9 ab$
$ab = 18$
Similarly, $(a+b)^2 = (a -b)^2  + 4ab$
$(a+b)^2 = 3^2 + 4(18)$
$(a+b)^2 = 9 + 72 $
$a +b = \pm 9$
The general form of equation is $x^2 -Sx + P = 0 $, hence the equation will be
$x^2 \pm 9x + 18 = 0$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\alpha , \beta$ are the roots of the equation $ { x }^{ 2 } - 2x + 3 = 0$, obtain the equation whose roots are ${ \alpha  }^{ 3 } - 3{ \alpha  }^{ 2 } + 5\alpha - 2,  { \beta  }^{ 3 } - { \beta  }^{ 2 } + \beta + 5$.

  1. ${x}^{2}-3x+2=0$
  2. ${x}^{2}+3x-2=0$
  3. $-{x}^{2}-3x+2=0$
  4. $-{x}^{2}+3x-2=0$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

If $\alpha, \beta$ are the roots of $x^2-2x+3=0$
then $\displaystyle \alpha ^{2}-2\alpha +3= 0$ ...(1)
and $\displaystyle \beta^2-2\beta+3=0$  ....(2)
$\displaystyle \therefore \alpha ^{2}= 2\alpha -3, \alpha ^{3}= 2\alpha ^{2}-3\alpha $
$\displaystyle \therefore P= \left ( 2\alpha ^{2}-3\alpha  \right )-3\alpha ^{2}+5\alpha -2$
$\displaystyle = -\alpha ^{2}+2\alpha -2= 3-2= 1,$ by (1)
Similarly $\displaystyle Q= 2 \therefore S= 3, P= 2$
Hence reqd. eq. is $\displaystyle x^{2}-3x+2= 0.$ or $-x^2+3x-2=0$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If the difference of the roots of a quadratic equation is 4 and the difference of their cubes is 208, then the quadratic equation is $x^{2}\, \pm\, 8x\, +\, 12\, =\, 0$
State true or false.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the roots of the equation be a and b
then, $a^3 - b^3 = 208$
and $a - b = 4$
cubing both sides:
$(a-b)^3 = 64 $
$a^3 - b^3 - 3ab (a-b) = 64$
$208 -3ab(4) = 64 $
$144 = 12 ab$
$ab = 12$
Similarly, $(a+b)^2 = (a -b)^2  + 4ab$
$(a+b)^2 = 4^2 + 4(12)$
$(a+b)^2 = 16 + 48 $
$a +b = \pm 8$
The general form of equation is $x^2 -Sx + P = 0 $, hence the equation will be
$x^2 \pm 8x + 12 = 0$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\alpha$ and $\beta$ are the roots of the equation $4x^{2}\, -\, 5x\, +\, 2\, =\, 0$, find the equation whose roots are
$\alpha\, +\, \displaystyle \frac{1}{\alpha}$ and $\beta\, +\, \displaystyle \frac{1}{\beta}$.

  1. $8x^{2}\, +\, 30x\, +\, 29\, =\, 0$
  2. $x^{2}\, -\, 30x\, +\, 29\, =\, 0$
  3. $8x^{2}\, -\, 30x\, +\, 29\, =\, 0$
  4. $x^{2}\, +\, 30x\, +\, 29\, =\, 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation is: $4x^2 - 5x + 2 = 0 $


Sum of the roots = $\dfrac{5}{4}$

Product of the roots  = $\dfrac{2}{4} = \dfrac{1}{2}$

If the roots are $\alpha + \dfrac{1}{\alpha}, \beta + \dfrac{1}{\beta}$

Sum of roots = $\alpha + \dfrac{1}{\alpha} + \beta + \dfrac{1}{\beta}$
 
= $\alpha + \beta + \dfrac{\alpha+ \beta}{\alpha\beta}$

= $\dfrac{5}{4} + \dfrac{\dfrac{5}{4}}{\dfrac{1}{2}}$

= $\dfrac{5}{4} + \dfrac{5}{2}$

$\dfrac{15}{4}$

Product of roots = $(\alpha + \dfrac{1}{\alpha})( \beta + \dfrac{1}{\beta})$

= $\alpha\beta + \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} + \dfrac{1}{\alpha\beta}$

= $\dfrac{1}{2} + \dfrac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta} + 2$

= $\dfrac{1}{2} + \dfrac{\dfrac{25}{16} - 1}{\dfrac{1}{2}} + 2$

= $\dfrac{1}{2} + \dfrac{9}{8} + 2$

= $\dfrac{4 + 9 + 16}{8}$ 

= $\dfrac{29}{8}$

Hence.the equation in the standard form, $x^2 - Sx + P  = 0$ can be written as:

=$x^2 - \dfrac{15}{4}x + \dfrac{29}{8} = 0$

= $8x^2 - 30x + 29 = 0$

Hence option $'C'$ is the answer.

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\alpha$ and $\beta$ are the roots of the equation $4x^{2}\, -\, 5x\, +\, 2\, =\, 0$, find the equation whose roots are
$\displaystyle \frac{\alpha}{\beta}$ and $\displaystyle \frac{\beta}{\alpha}$.

  1. $8x^{2}\, +\, 9x\, +\, 8\, =\, 0$
  2. $8x^{2}\, -\, 9x\, +\, 8\, =\, 0$
  3. $8x^{2}\, -\, 9x\, -\, 8\, =\, 0$
  4. $x^{2}\, -\, 9x\, +\, 8\, =\, 0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$4x^2 - 5x + 2 = 0 $


If $\alpha$ and $\beta$ are the roots of this equation, 

then , sum of roots: $\alpha + \beta$ = $\dfrac{5}{4}$

Product of roots: $\alpha. \beta = \dfrac{2}{4}$

The equation which has roots as : $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$

Sum of roots: $\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha}$ 

= $\dfrac{\alpha^2 + \beta^2}{\alpha\beta}$

= $\dfrac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}$

= $\dfrac{ \left (\dfrac{5}{4} \right )^2 - 2\dfrac{2}{4}}{\dfrac{2}{4}}$

= $\dfrac{9}{8}$

Product of roots: $ \left (\dfrac{\alpha}{\beta} \right ) \left (\dfrac{\beta}{\alpha} \right )= 1$

Thus new equation is :$x^2 -Sx + P = 0$

$x^2 - \dfrac{9}{8} + 1= 0$

$8x^2 - 9x + 8 = 0$