Question

If the zeroes of the quadratic polynomial ${\mathrm{x}}^2+(\mathrm{a}+1)\mathrm{x}+\mathrm{b}$ are $2$ and $-3$, then

• A. $\mathrm{a}=-7$, $\mathrm{b}=-1$
• B. $\mathrm{a}=5$, $\mathrm{b}=-1$
• C. $\mathrm{a}=2$, $\mathrm{b}=-6$
• D. $\mathrm{a}=0$, $\mathrm{b}=-6$

Answer 1

The correct option is D

$\mathrm{a}=0$, $\mathrm{b}=-6$

Explanation for correct option

Step 1 General equation of the Quadratic polynomials

As we all know that, if $\mathrm{\alpha }$ and $\mathrm{\beta }$ are the zeroes of polynomial ${\mathbf{ax}}^{\mathbf{2}}\mathbf{+}\mathbf{bx}\mathbf{+}\mathbf{c}$ then,

Sum of the roots are,

$\mathbf{\alpha }\mathbf{+}\mathbf{\beta }\mathbf{=}\frac{\mathbf{-}\mathbf{b}}{\mathbf{a}}$

Here, $\mathrm{a}$ is the coefficient of ${\mathrm{x}}^3$ and $\mathrm{b}$ is the coefficient of ${\mathrm{x}}^2$.

Product of the roots are,

$\mathbf{\alpha }\mathbf{\times }\mathbf{\beta }\mathbf{=}\frac{\mathbf{-}\mathbf{d}}{\mathbf{a}}$

Here, $\mathrm{a}$ is the coefficient of ${\mathrm{x}}^3$ and $\mathrm{d}$ is the constant term.

It is given that the polynomial equation is ${\mathrm{x}}^2+(\mathrm{a}+1)\mathrm{x}+\mathrm{b}$ and the zeroes are $2$ and $-3$.

We will find the value of $\mathrm{a}$ by using the formula of sum of roots and value of $\mathrm{b}$ by using the formula of product of roots.

Step 2. Find the value of $\mathrm{a}$.

From the given, the zeroes are $2$ and $-3$.

Let,

$\mathrm{\alpha }=2$

$\mathrm{\beta }=-3$

Put the values in the formula for sum of zeroes, we get,

$\Rightarrow$ $2+\left(-3\right)=\frac{-\left(\mathrm{a}+1\right)}{1}$

$\Rightarrow$ $-1=-\left(\mathrm{a}+1\right)$

$\Rightarrow$ $1=\left(\mathrm{a}+1\right)$

$\Rightarrow$ $\mathrm{a}=0$

Step 3. Find the value of $\mathrm{b}$.

Use product of roots formula to find the value of $\mathrm{b}$, we get

$\Rightarrow$ $2\times \left(-3\right)=\frac{\mathrm{b}}{1}$

$\Rightarrow$ $-6=\mathrm{b}$

$\Rightarrow$ $\mathrm{b}=-6$

Hence, the value of $\mathrm{a}$ is $0$ and value of $\mathrm{b}$ is $-6$.

Therefore, option $\mathrm{D}$ is correct answer.