Tag: forming quadratic equation

Questions Related to forming quadratic equation

If $\alpha , \beta$ are the roots of the equation $ax^2+bx+c=0$ then the quadratic equation whose roots are $\alpha + \beta , \alpha \beta$ is:

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

  1. $a^2 x^2 +a(b-c) x+bc=0$

  2. $a^2 x^2 + a(b-c) x-bc=0$

  3. $ax^2 +(b+c) x+bc=0$

  4. $ax^2-(b+c)x-bc=0$

Show answer
Correct Option: B
Explanation:

From the first equation, we can conclude that
$\alpha+\beta=-\cfrac{b}a$  ...(i)

$\alpha\beta=\cfrac{c}a$      ...(ii)

Therefore, the new equation will be

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

Substituting the values from (i) and (ii), we get

$x^2-\left(\cfrac{-b+c}{a}\right)x+\left(\cfrac{-bc}{a^2}\right)=0$

$a^2x^2+a(b-c)x-bc=0$


Hence, the answer is
$a^2x^2+a(b-c)x-bc=0$

If $\alpha$ and $\beta$ are the roots of the equation $ax^2+bx+c=0$ and if $px^2+qx+r=0$ has roots $\displaystyle \frac{1-\alpha}{\alpha}$ and $\displaystyle \frac{1-\beta}{\beta}$, then $r$ is

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

  1. $a+2b$

  2. $a+b+c$

  3. $ab+bc+ca$

  4. $abc$

Show answer
Correct Option: B
Explanation:
The equation with roots $\cfrac1{\alpha}$ and $\cfrac1{\beta}$
$=a\left(\cfrac1x\right)^2+b\left(\cfrac1x\right)+c$
$=cx^2+bx+a=0$ ..(1)
Now $\cfrac{1-\alpha}{\alpha}=\cfrac{1}{\alpha}-1$
Similarly $\cfrac{1-\beta}{\beta}=\cfrac{1}{\beta}-1$
Therefore the quadratic equation containing these roots is
$c\left(x+1\right)^2+b\left(x+1\right)+a$
$=cx^2+\left(b+2c\right)x+a+b+c = 0$
By comparing coefficients we get with $px^2+qx+r=0$ we get
$r=a+b+c$

If $\alpha , \beta$ are the roots of the equation $9x^2+6x+1=0$, then the equation with the roots $\cfrac{1}{\alpha}, \cfrac{1}{\beta}$ is :

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

  1. $2x^2+3x+18=0$

  2. $x^2+6x-9=0$

  3. $x^2+6x+9=0$

  4. $x^2-6x+9=0$

Show answer
Correct Option: C
Explanation:
$\left(x-\cfrac1{\alpha}\right)\left(x-\cfrac1{\beta}\right)=0$
$x^2-\left(\cfrac{1}{\alpha}+\cfrac{1}{\beta}\right)x+\left(\cfrac{1}{\alpha\beta}\right)=0$
$x^2-\left(\cfrac{\alpha+\beta}{\alpha\beta}\right)x+\left(\cfrac{1}{\alpha\beta}\right)=0$
From the given equation we know
$\alpha+\beta=-\cfrac69$
$\alpha\beta=\cfrac19$
By substituting we get
$x^2-\left(-6\right)x+9=0$
$x^2+6x+9=0$

If $\alpha$ and $\beta$ are roots of $2{ x }^{ 2 }-3x-6=0$, then the equation whose roots are ${ \alpha  }^{ 2 }+2$ and ${ \beta  }^{ 2 }+2$ will be

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

  1. $4{ x }^{ 2 }+49x-118=0$

  2. $4{ x }^{ 2 }-49x-118=0$

  3. $4{ x }^{ 2 }-49x+118=0$

  4. $4{ x }^{ 2 }+49x+118=0$

Show answer
Correct Option: C
Explanation:

$2x^{2}-3x-6=0$

$\alpha+\beta=\dfrac{3}{2}$
$\alpha\beta=-3$
Now roots are $\alpha^{2}+2$   and   $\beta^{2}+2$
$sum=\alpha^{2}+2+\beta^{2}+2$
$=(\alpha+\beta)^{2}-2\alpha\beta+4$
$=\dfrac{9}{4}+6+4$
$=\dfrac{49}{4}$
$Product=(\alpha^{2}+2)(\beta^{2}+2)$
$=\alpha^{2}\beta^{2}+2(\alpha^{2}+\beta^{2})+4$
$=9+2\times\dfrac{33}{4}+4=13+\dfrac{33}{2}=\dfrac{59}{2}$
Equation
$x^{2}-\dfrac{49x}{4}+\dfrac{59}{2}=0$
$4x^{2}-49x+118=0$

If $\alpha, \beta$ are the roots of $x^2 + px+1=0$ and $\gamma, \delta $ are the roots of $x^2+qx+1=0$, then $(\alpha - \gamma) (\beta - \gamma)(\alpha - \delta) (\beta + \delta)=$

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

  1. $2q^2$

  2. $2p^2$

  3. $p^2-q^2$

  4. $q^2 - p^2$

Show answer
Correct Option: D
Explanation:

$\alpha, \beta$ are the roots of $x^2 + px+1=0$
$\Rightarrow \alpha+\beta = -p,  \alpha \beta =1$

$\gamma, \delta$ are the roots of $x^2+qx+1=0$
$\Rightarrow \gamma \delta =1, \gamma^2+q\gamma +1=0$,
$\delta^2 +q\delta +1=0$

$(\alpha - \gamma)(\beta -\gamma)(\alpha + \delta)(\beta + \delta)$
$=[\alpha \beta - \gamma (\alpha + \beta)+\gamma^2][\alpha \beta + \delta (\alpha + \beta) + \delta^2]$
$=(1+p \gamma + \gamma^2)(1-p\delta + \delta^2)$
$=(p \gamma - q \gamma)(-p \delta - q\delta)$
$=-\gamma \delta (p-q)(p+q)$
$=-(p^2-q^2) = q^2 - p^2$

Hence, option D.

Find the equation whose sum of roots and product of roots are the product and sum of roots of $x^2 + 5x + 6 = 0$ respectively.

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

  1. $x^2 - 6x - 5 = 0$

  2. $x^2 - 5x - 6 = 0$

  3. $x^2 + 11x - 1 = 0$

  4. None of the above

Show answer
Correct Option: A
Explanation:

In the given equation
sum of roots $= -5$ and product of roots $= 6$
The standard form of a quadratic equation is: $x^2 - (S)x + P = 0$, where S and P are sum and product of roots.
So according to question
$x^2 - (6)x + (-5) = 0$
The required equation will be $x^2 - 6x - 5 = 0$

If $\alpha, \beta $ are the roots of $ax^2+bx+c=0$ then the equation whose roots are $2+\alpha , 2+\beta$ is:

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

  1. $ax^2+x(4a-b) + 4a-2b+c=0$

  2. $ax^2+x(4a-b) + 4a+2b+c=0$

  3. $ax^2+x(b-4a) = 4a+2b+c=0$

  4. $ax^2+x(b-4a) + 4a-2b+c=0$

Show answer
Correct Option: D
Explanation:

$\alpha, \beta$ are the roots of $\Rightarrow { ax }^{ 2 }+bx+c=0$
Then, $a(\alpha)^2 + b(\alpha)+c =0$
Now, $\alpha +2 = x $
Hence, $\alpha = x -2$
Thus, replace $\alpha$ by $x-2$ in the given equation,
Required equation is
$a(x-2)^2+b(x-2)+c=0$
$\Rightarrow a(x^2-4x+4)+bx-2b+c=0$
$\Rightarrow ax^2 +(b-4a) x+(4a-2b+c)=0$
$\Rightarrow ax^{ 2 }+x(b-4a)+4a-2b+c=0$

If $\alpha , \beta$ are the roots of the equation $x^2 - 3x + 1 = 0$, then the equation with roots $\displaystyle \frac{1}{\alpha - 2} , \frac{1}{\beta - 2}$ will be

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

  1. $x^2- x- 1 = 0$

  2. $x^2 + x - 1 = 0$

  3. $x^2 + x + 2 = 0$

  4. none of these

Show answer
Correct Option: A
Explanation:
$\alpha , \beta$ are the roots of the equation $x^2 - 3x + 1 = 0$
$\Rightarrow  \alpha^2-3\alpha+1=0$ ------(1)
Let $\displaystyle\dfrac{1}{\alpha-2}=y$
$\Rightarrow\displaystyle \alpha=2+\dfrac{1}{y}$
From (1), we get
$\displaystyle\left(2+\dfrac{1}{y}\right)^2-3\left(2+\dfrac{1}{y}\right)+1=0$
$\Rightarrow\displaystyle \dfrac{(2y+1)^2}{y^2}-\dfrac{3(2y+1)}{y}+1=0$
$\Rightarrow y^2-y-1=0$
$\therefore$ The equation with roots $\displaystyle \dfrac{1}{\alpha - 2} , \dfrac{1}{\beta - 2}$ is $x^2-x-1=0$
Hence, option A.

If $\alpha, \beta$ are roots of $ax^2+bx+c=0$, then one root of the equation $ax^2-bx(x-1) + c(x-1)^2=0$ is :

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

  1. $\displaystyle \left ( \frac{\alpha}{1- \alpha} \right )$

  2. $\displaystyle \left ( \frac{1-\beta}{\beta} \right )$

  3. $\displaystyle \left ( \frac{\alpha}{1+ \alpha} \right )$

  4. $\displaystyle \left ( \frac{\beta}{1+ \beta} \right )$

Show answer
Correct Option: C,D
Explanation:

We have, $ax^2-bx^2+bx+cx^2-2cx+c=0$

$(a-b+c)x^2+(b-2c)x+c=0$

Sum of the roots (S)
$\displaystyle \frac{b-2c}{a-b+c} = \frac{\left ( -\frac{b}{a} + \frac{2c}{a} \right )}{\left ( 1- \frac{b}{a} + \frac{c}{a}\right )}$

$\displaystyle S = \frac{\alpha+\beta+1\alpha \beta}{2+ \alpha + \beta+ \alpha \beta}=\frac{\alpha}{\alpha+1}+ \frac{\beta}{\beta+1}$

Product of the roots (P) $\displaystyle =\frac{c}{a-b+c}$

$\Rightarrow \displaystyle P= \frac{\left ( \frac{c}{a} \right )}{\left ( 1- \frac{b}{c}+\frac{c}{a}\right )}$

$\displaystyle=\frac{\alpha \beta}{1+\alpha+\beta+\alpha \beta} = \frac{\alpha}{(\alpha+1)} \cdot \frac{\beta}{(\beta+1)}$

Thus the roots are $ \displaystyle \frac{\alpha}{\alpha+1} and \frac{\beta}{\beta+1}$

If $\alpha $ and $\beta$ be the roots of the equation $x^{2}+px+q = 0$, then the equation whose roots are $\alpha^{2}+\alpha\beta$ and $\beta^{2}+\alpha\beta$ is

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

  1. $x^{2}+p^{2}x+p^{2}q = 0$

  2. $x^{2}-q^{2}x+p^{2}q = 0$

  3. $x^{2}+q^{2}x+p^{2}q = 0$

  4. $x^{2}-p^{2}x+p^{2}q = 0$

Show answer
Correct Option: D
Explanation:

Since $\alpha$ and $\beta$ are roots of the equation
$x^{2}+px+q = 0$, therefore
$\alpha + \beta = -p$      ...(i)
and $\alpha\beta = q$      ...(ii)
Sum of the roots $= \alpha^{2}+\alpha\beta+\beta^{2}+\alpha\beta$
$= (\alpha+\beta)^{2} = p^{2}$
Product of the roots $=(\alpha^{2}+\alpha\beta)(\beta^{2}+\alpha\beta)$
$= \alpha\beta (\alpha+\beta)^{2} = qp^{2}$
Required equation will be
$x^{2}$-(Sum  of  the  roots)$x$ + Product  of  the  roots = $0$
or $x^{2}-p^{2}x+qp^{2} = 0$