Question

If the squared difference of the zeros of the quadratic polynomial $f\left( x \right) = {{x}^{2}}+px+45$ is equal to 144, find the value of p.

Answer 1

Hint: We will be using the concept of quadratic equations to solve the problem. We will be using sum of roots and product of roots to further simplify the problem.

Complete Step-by-Step solution:
Now, we have a quadratic expression as $f\left( x \right)={{x}^{2}}+px+45$ and we have been given the squared difference of the zeros of the polynomial as 144.
Now, we have to find the value of P.
Now, to find this value we will be using the sum of roots and product of roots. We know that in a quadratic equation $a{{x}^{2}}+bx+c=0$.
$\begin{align}
  & \text{sum of roots }=\dfrac{-b}{a} \\
 & \text{product of roots }=\dfrac{c}{a} \\
\end{align}$
Therefore, in$f\left( x \right)={{x}^{2}}+px+45$.
$\begin{align}
  & \alpha +\beta =\text{sum of roots }=-p..........\left( 1 \right) \\
 & \alpha \beta \ \text{=}\ \text{product of roots }=45...........\left( 2 \right) \\
\end{align}$
Now, we have been given that the square of the difference of the roots is 144. Therefore,
${{\left( \alpha -\beta \right)}^{2}}=144$
Now, we know that,
$\begin{align}
  & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
 & {{\alpha }^{2}}+{{\beta }^{2}}-2\alpha \beta =144 \\
\end{align}$
Now, we will add and subtract $2\alpha \beta $ in the equation so we have,
$\begin{align}
  & {{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta -4\alpha \beta =144 \\
 & {{\left( \alpha +\beta \right)}^{2}}-4\alpha \beta =144 \\
 & {{\left( -p \right)}^{2}}-4\left( 4s \right)=144 \\
 & {{p}^{2}}=144+4\left( 4s \right) \\
 & {{p}^{2}}=144+180 \\
 & {{p}^{2}}=324 \\
 & p=\pm \sqrt{324} \\
 & p=\pm 18 \\
\end{align}$
Therefore, the value of p is $\pm 18$.

Note: To solve these types of questions one must know how to find the relation between sum of roots, product of roots and coefficient of quadratic equation.
$\begin{align}
  & \text{sum of roots }=\dfrac{-b}{a} \\
 & \text{product of roots }=\dfrac{c}{a} \\
\end{align}$