Question

If the squared difference of the zeroes of x² + px + 45 is equal to 216, find the value of p .​

Answer 1

ANSWER:

Given:

• p(x) = x² + px + 45
• (α - β)² = 216

To Find:

• Value of p

Solution:

We have a polynomial,

$:\implies p(x)=x^2+px+45$

Let, the zeroes of the polynomial be α and β.

We are given that,

$:\implies (\alpha-\beta)^2=216$

We know that,

⇒ (a - b)² = a² + b² - 2ab

So,

$:\implies (\alpha)^2+(\beta)^2-2\alpha\beta=216$

Now, We know that,

⇒ a² + b² = (a + b)² - 2ab

So,

$:\implies (\alpha+\beta)^2-2\alpha\beta-2\alpha\beta=216$

Hence,

$:\implies (\alpha+\beta)^2-4\alpha\beta=216$

Now, for a quadratic equation, ax² + bx + c = 0,

⇒ Sum of zeroes = -b/a

And,

⇒ Product of zeroes = c/a

So,

$:\implies (\alpha+\beta)^2-4(\alpha\beta)=216$

$:\implies \left(\dfrac{-(p)}{1}\right)^2-4\left(\dfrac{45}{1}\right)=216$

Hence,

$:\implies (-p)^2-4(45)=216$

Solving it,

$:\implies p^2-180=216$

Transposing 180 to RHS,

$:\implies p^2=216+180$

$:\implies p^2=396$

So,

$:\implies p=\pm\sqrt{396}$

Hence,

$:\implies\bf p=\pm6\sqrt{11}$