Transformation of Quadratic Equations

Quadratic Equation:

A polynomial equation in which the highest degree of a variable term is 2 is called a quadratic equation.
Standard form of quadratic equation is $a x^2+b x+c=0$
Where $\mathrm{a}, \mathrm{b}$, and c are constants (they may be real or imaginary) and called the coefficients of the equation and $a \neq 0$ ( a is also called the leading coefficient).

$
\text { Eg, }-5 x^2-3 x+2=0, x^2=0,(1+i) x^2-3 x+2 i=0
$

As the degree of the quadratic polynomial is 2 , so it always has 2 roots (number of real roots + number of imaginary roots $=2$ )

Roots of quadratic equation

The root of the quadratic equation is given by the formula:

$
\begin{aligned}
& \mathrm{x}=\frac{-\mathrm{b} \pm \sqrt{\mathrm{D}}}{2 \mathrm{a}} \\
& \text { or } \\
& \mathrm{x}=\frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^2-4 \mathrm{ac}}}{2 \mathrm{a}}
\end{aligned}
$

Where $D$ is called the discriminant of the quadratic equation, given by $D=b^2-4 a c$, The discriminant of a quadratic equation reveals the nature of roots.

Equation from roots

A quadratic equation with $\alpha$ and $\beta$ as its roots is $(\mathrm{x}-\alpha)(\mathrm{x}-\beta)=0$

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

So, the equation with given roots can be written as $x^2-$ (sum of roots) $x+($ Product of roots) $=0$

Transformation of roots

Let $\alpha$ and $\beta$ be the roots of quadratic equation $\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0$, then
i) the equation with root $\alpha+\mathrm{k}$ and $\beta+\mathrm{k}$ will be

$
a(x-k)^2+b(x-k)+c=0, \quad(\text { replace } \mathrm{x} \text { by } \mathrm{x}-\mathrm{k})
$

ii) the equation with root $\alpha-k$ and $\beta-k$ will be

$
\left.a(x+k)^2+b(x+k)+c=0, \text { (replace } \mathrm{x} \text { by } \mathrm{x}+\mathrm{k}\right)
$

iii) the equation with root $\alpha$ k and $\beta \mathrm{k}$ will be

$
a x^2+k b x+k^2 c=0\left(\text { replace } x \text { by } \frac{x}{k}\right)
$

iv) the equation with root $\frac{\alpha}{k}$ and $\frac{\beta}{k}$ will be $a x^2 k^2+b k x+c=0$ (replace x by kx )
v) the equation with root $\alpha$ and $-\beta$ will be $a x^2-b x+c=0 \quad$ (replace x by -x )
vi) the equation with root $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ will be

$
c x^2+b x+a=0\left(\text { replace } \times \text { by } \frac{1}{x}\right)
$

vii) the equation with root $-\frac{1}{\alpha}$ and $-\frac{1}{\beta}$ will be $c x^2-b x+a=0\left(\right.$ replace $\times$ by $\left.-\frac{1}{x}\right)$
viii) the equation with root $\frac{k}{\alpha}$ and $\frac{k}{\beta}$ will be

$
c x^2+k b x+k^2 a=0\left(\text { replace } \times \text { by } \frac{k}{x}\right)
$

Recommend Video Based on Transformation of Quadratic Equations:

Solved Examples Based on Transformation of Quadratic Function:

Example 1: Let $\alpha, \beta, \gamma$ are roots of $x^3-x^2+1=0$ then the equation whose roots are $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$ is
1) $x^3+x-1=0$
2) $x^3-x+1=0$
3) $x^3+x+1=0$
4) $x^3-x-1=0$

Solution

As we learned in

Transformation of the equation -

To find an equation whose roots are symmetrical functions of $\alpha$ and $\beta$, Where $\alpha \& \beta$ are roots of some other equation.
- wherein

Take any of the roots to be equal to $y$ \& calculate $\alpha$ or $\beta$ accordingly in terms of $y$ \& satisfy the given equation to get required equation.

Let $\frac{1}{\alpha}=y \Rightarrow \alpha=\frac{1}{y}$
Now, $\alpha$ being the root of a given equation will satisfy it, So

$
\begin{aligned}
& \left(\frac{1}{y}\right)^3-\left(\frac{1}{y}\right)^2+1=0 \Rightarrow 1-y+y^3=0 \\
& \Rightarrow y^3-y+1=0
\end{aligned}
$

$\therefore$ equation is $x^3-x+1=0$
Hence, the answer is the option (2).
Example 2: Let $\alpha, \beta, \gamma$ are roots of $x^3+x+1=0$ then the equation with roots $\alpha-1, \beta-1, \gamma-1$ is :
1) $x^3+3 x^2+4 x+3=0$
2) $x^3-3 x^2+4 x+3=0$
3) $x^3-3 x^2-4 x+3=0$
4) $x^3-3 x^2+4 x-3=0$

Solution

As we have learned

Transformation of the equation -

To find an equation whose roots are symmetrical functions of $\alpha$ and $\beta$, Where $\alpha \& \beta$ are roots of some other equation.
- wherein

Take any of the roots to be equal to $y \&$ calculate $\alpha$ or $\beta$ accordingly in terms of $y$ \& satisfy the given equation to get the required equation.

Let $\alpha-1=y \Rightarrow \alpha=y+1$, Now put in the given equation we get -

$
(y+1)^3+(y+1)+1=0 \Rightarrow y^3+3 y^2+4 y+3=0
$

$\therefore$ Equation is $\rightarrow x^3+3 x^2+4 x+3=0$
Hence, the answer is the option (1).

Example 3: If $\alpha, \beta, \gamma$ are roots of $x^3-x+2=0$ then the equation whose roots are $\alpha+\beta+2 \gamma, \beta+\gamma+2 \alpha \& \gamma+\alpha+2 \beta$ will be:
1) $x^3+x-2=0$
2) $x^3+x+2=0$
3) $x^3-x+2=0$
4) $x^3-x-2=0$

Solution

As we learned in

Transformation of the equation -

To find an equation whose roots are symmetrical functions of $\alpha$ and $\beta$, Where $\alpha \& \beta$ are roots of some other equation.

- wherein

Let $y=\alpha+\beta+2 \gamma=\alpha+\beta+\gamma+\gamma=0+\gamma$
$\Rightarrow \gamma=y$
now, $\gamma$ will satisfy the given equation so $y^3-y+2=0$
$\therefore$ equation is $x^3-x+2=0$
Hence, the answer is the option (3).

Example 4: If $\alpha$ and $\beta$ are the roots of the equation $x^2+p x+2=0$ and $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ are the roots of the equation $2 x^2+2 p x+1=0$, then $\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right)\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)$ is equal to:
1) $\frac{9}{4}\left(9+q^2\right)$
2) $\frac{9}{4}\left(9-q^2\right)$
3) $\frac{9}{4}\left(9+p^2\right)$
4) $\frac{9}{4}\left(9-p^2\right)$

Solution

$
\begin{aligned}
& x^2+p x+2=0 \quad \alpha, \beta \\
& 2 x^2+2 p x+1=0 \quad \frac{1}{\alpha}, \frac{1}{\beta}
\end{aligned}
$

$
\begin{aligned}
& \alpha+\beta=-p, \alpha \beta=2 \\
& \frac{1}{\alpha}+\frac{1}{\beta}=-q, \frac{1}{\alpha \beta}=\frac{1}{2} \\
& \left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right)\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right) \\
& =\left(\alpha \beta-\frac{\alpha}{\beta}-\frac{\beta}{\alpha}+\frac{1}{\alpha \beta}\right)\left(\alpha \beta+2+\frac{1}{\alpha \beta}\right) \\
& =\left(\frac{5}{2}-\frac{\alpha^2+\beta^2}{\alpha \beta}\right)\left(\frac{9}{2}\right) \\
& =\left(\frac{5}{2}-\frac{p^2-4^{-1}}{2}\right)\left(\frac{9}{2}\right) \\
& =\frac{9}{4}\left(9-p^2\right)
\end{aligned}
$
Hence, the answer is the option 4.

Example 5: The number of roots of the equation, $(81)^{\sin ^2 x}+(81)^{\cos ^2 x}=30$ in the interval $[0, \pi]$ is equal to :
1) $4$
2) $2$
3) $8$
4) $3$

Solution
$
\begin{aligned}
& (81)^{\sin ^2 x}+(81)^{\cos ^2 x}=30 \\
& (81)^{\sin ^2 x}+\frac{(81)^1}{(18)^{\sin ^2 x}}=30 \\
& (81)^{\sin ^2 x}=t \\
& t+\frac{81}{t}=30 \\
& t^2-30 t+81=0 \\
& (t-3)(t-27)=0 \\
& (81)^{\sin ^2 x}=3^1 \quad \text { or } \quad(81)^{\sin ^2 x}=3^3 \\
& 3^{4 \sin ^2 x}=3^1 \quad \text { or } \quad 3^{4 \sin ^2 x}=3^3 \\
& \sin ^2 x=\frac{1}{4} \quad \text { or } \quad \sin ^2 x=\frac{3}{4}
\end{aligned}
$