If the squared difference of the zeros of the quadratic polynomial f(x)
Question:

If the squared difference of the zeros of the quadratic polynomial $1(x)=x^{2}+p x+45$ is equal to 144, find the value of $p$.

Solution:

Given $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f(x)=x^{2}+p x+45$

$\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

$=\frac{-p}{1}$

= -p

$\alpha \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

$=\frac{45}{1}$

=45

We have, 

$(\alpha-\beta)^{2}=\alpha^{2}+\beta^{2}-2 \alpha \beta$

$144=(\alpha+\beta)^{2}-2 \alpha \beta-2 \alpha \beta$

$144=(\alpha+\beta)^{2}-4 \alpha \beta$

Substituting $\alpha+\beta=-p$ and $\alpha \beta=45$ then we get,

$144=(-p)^{2}-4 \times 4$

$144=p^{2}-4 \times 45$

$144=n^{2}-180$

$144+180=p^{2}$

$324=p^{2}$

$\sqrt{18 \times 18}=p \times p$

$\pm 18=p$

Hence, the value of $p$ is $\pm 18$.

Administrator

Leave a comment

Please enter comment.
Please enter your name.