Question

A quadratic polynomial whose zeroes are -4 and 5 is,
(A) ${{x}^{2}}-3x-10$
(B) ${{x}^{2}}+x-20$
(C) ${{x}^{2}}-6x-40$
(D) None of these

Answer 1

Hint: We solve this question by first considering the quadratic equation, $a{{x}^{2}}+bx+c$, with the roots $\alpha ,\beta $. Then we use the formula for the sum of the roots, $\alpha +\beta =-\dfrac{b}{a}$ and formula for the product of the roots, $\alpha \beta =\dfrac{c}{a}$. Then we substitute the given values of roots in the above formulas and find the relation between $b$ and $a$, $c$ and $a$ respectively. Then we substitute these relations in the quadratic equation $a{{x}^{2}}+bx+c$ and then take $a$ common from it and then find the required equation.

Complete step-by-step answer:
We are given that the zeroes of a quadratic polynomial are -4 and 5.
Now let us consider the formula for the sum and product of roots. For any quadratic polynomial, $a{{x}^{2}}+bx+c$, if the roots of the polynomial are $\alpha ,\beta $, then we can write the sum and the product of the roots of the equation as.
Sum of the roots = $\alpha +\beta =-\dfrac{b}{a}$
Product of the roots = $\alpha \beta =\dfrac{c}{a}$
So, let us substitute the given values in the sum of the roots in the above formula. Then we get,
$\begin{align}
  & \Rightarrow -4+5=-\dfrac{b}{a} \\
 & \Rightarrow 1=-\dfrac{b}{a} \\
 & \Rightarrow a=-b \\
 & \Rightarrow b=-a.........\left( 1 \right) \\
\end{align}$
So, let us substitute the given values in the product of the roots in the above formula. Then we get,
$\begin{align}
  & \Rightarrow -4\times 5=\dfrac{c}{a} \\
 & \Rightarrow -20=\dfrac{c}{a} \\
 & \Rightarrow -20a=c \\
 & \Rightarrow c=-20a..........\left( 2 \right) \\
\end{align}$
As the quadratic equation is $a{{x}^{2}}+bx+c$, using equations (1) and (2) we can write it as,
$\begin{align}
  & \Rightarrow a{{x}^{2}}+bx+c=0 \\
 & \Rightarrow a{{x}^{2}}-ax-20a=0 \\
 & \Rightarrow a\left( {{x}^{2}}-x-20 \right)=0 \\
 & \Rightarrow {{x}^{2}}-x-20=0 \\
\end{align}$
Hence the required quadratic equation is ${{x}^{2}}-x-20$.
Hence the answer is Option D.

So, the correct answer is “Option D”.

Note: We can also solve this question in an alternative method.
We can solve this question by substituting the given values of roots in the given options and see if they satisfy them.
We are given that the roots are -4 and 5.
Now let us substitute these values in the given options.
(A) ${{x}^{2}}-3x-10$
Let us substitute the value $x=-4$ in it. Then we get,
$\begin{align}
  & \Rightarrow {{\left( -4 \right)}^{2}}-3\left( -4 \right)-10 \\
 & \Rightarrow 16-\left( -12 \right)-10 \\
 & \Rightarrow 16+12-10 \\
 & \Rightarrow 18 \\
\end{align}$
So, -4 is not a root of this equation. So, this option is not correct.

(B) ${{x}^{2}}+x-20$
Let us substitute the value $x=-4$ in it. Then we get,
$\begin{align}
  & \Rightarrow {{\left( -4 \right)}^{2}}+\left( -4 \right)-20 \\
 & \Rightarrow 16+\left( -4 \right)-20 \\
 & \Rightarrow 16-4-20 \\
 & \Rightarrow -8 \\
\end{align}$
So, -4 is not a root of this equation. So, this option is not correct.

(C) ${{x}^{2}}-6x-40$
Let us substitute the value $x=-4$ in it. Then we get,
$\begin{align}
  & \Rightarrow {{\left( -4 \right)}^{2}}-6\left( -4 \right)-40 \\
 & \Rightarrow 16-\left( -24 \right)-40 \\
 & \Rightarrow 16+24-40 \\
 & \Rightarrow 40-40 \\
 & \Rightarrow 0 \\
\end{align}$
So, -4 is a root of this equation.
Let us substitute the value $x=5$ in it. Then we get,
$\begin{align}
  & \Rightarrow {{\left( 5 \right)}^{2}}-6\left( 5 \right)-40 \\
 & \Rightarrow 25-\left( 30 \right)-40 \\
 & \Rightarrow 25-30-40 \\
 & \Rightarrow -5-40 \\
 & \Rightarrow -45 \\
\end{align}$
So, 5 is not a root of this equation. So, this option is not correct.
As all the options above are wrong, the answer is none of these.
Hence the answer is Option D.