Transformation of Quadratic Equations

Quadratic Equation plays an important role in mathematics due to their parabolic graphs and numerous applications. Transformation of these equations can help analysts to analyze different insights. The transformation of the parabolic equation includes shape, size, etc. It is useful in real-life applications.

This Story also Contains

1. Quadratic Equation:
2. Transformation of roots
3. Solved Examples Based on Transformation of Quadratic Function:

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}
$