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: Here zeros of the quadratic polynomial means roots of the quadratic polynomial i.e. the value should satisfy the given polynomial. Form a relation between
roots and p using the given condition.

Complete step-by-step answer:
Let $\alpha ,\beta $ be the roots or zeros of the quadratic polynomial $f\left( x \right) = {x^2} + px + 45$.
Given ${\left( {\alpha - \beta } \right)^2} = 144$
We know that for the quadratic polynomial \[a{x^2} + bx + c = 0\], the sum of the roots is \[ - \dfrac{b}{a}\] and the product of the roots is \[\dfrac{c}{a}\].
By using the above formula, for the quadratic polynomial $f\left( x \right) = {x^2} + px + 45$ ,
 the sum of the roots is equal to $\alpha + \beta = - p$
 the product of the roots is equal to $\alpha \beta = 45$
Now consider
$
  {\left( {\alpha - \beta } \right)^2} = 144 \\
  {\alpha ^2} + {\beta ^2} - 2\alpha \beta = 144 \\
$
Adding $4\alpha \beta $on both sides we get
\[
  {\alpha ^2} + {\beta ^2} - 2\alpha \beta + 4\alpha \beta = 144 + 4\alpha \beta \\
  {\alpha ^2} + {\beta ^2} + 2\alpha \beta = 144 + 4\alpha \beta \\
  {\left( {\alpha + \beta } \right)^2} = 144 + 4\alpha \beta \\
\]
Substituting the values $\alpha + \beta = - p$and $\alpha \beta = 45$ we get
$
  {\left( { - p} \right)^2} = 144 + 4\left( {45} \right) \\
  {p^2} = 144 + 180 \\
  {p^2} = 324 \\
$
Rooting on both sides we get,
$
  \sqrt {{p^2}} = \sqrt {324} \\
  p = \pm 18 \\
  \therefore p = \pm 18 \\
$

Note: In this problem quadratic polynomial means a polynomial with the degree (highest power) is \[2\]. The number of roots is always equal to the degree of the polynomial.