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

Given:

The squared difference of the zeros of the quadratic polynomial  $f(x)\ =\ x^2\ +\ px\ +\ 45$  is equal to 144.

To do:

Here, we have to find the value of $p$.

Solution: 

Let $α$ and $β$ be the zeros of the given quadratic polynomial.

 We know that, 

The standard form of a quadratic equation is $ax^2+bx+c=0$, where a, b and c are

constants and $a≠0$

Comparing the given equation with the standard form of a quadratic equation, 

$a=1$, $b=p$ and $c=45$

Sum of the roots $= α+β = \frac{-b}{a} = \frac{-p}{1} = -p$.

Product of the roots $= αβ = \frac{c}{a} = \frac{45}{1}=45$.

Therefore,

$(α-β)^2=144$

$(α+β)^2-4αβ=144$

$(-p)^2-4(45)=144$

$p^2-180-144=0$

$p^2-324=0$

$p^2=324$

$p=\sqrt{324}$

$p=18$ or $p=-18$

The value of $p$ is $-18$ or $18$.

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Updated on: 10-Oct-2022

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