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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