Question

If a, ß be the zeroes of the quadratic polynomial f (x) = x² + px + 48 and (a - B)² = 169, then p is equal​

Answer 1

we have given in the equation

f (x) = x² + px + 48

hence a+b = -p

αβ=45 also given that

$(a-b)^{2}$   =169

$\Rightarrow(\alpha+\beta)^{2}-4 \alpha \beta=169$

$(-\mathrm{p})^{2}-4 \times 48=169$

$\mathrm{p}^{2}-192=169$

$\mathrm{p}^{2}=169+192$

$\mathrm{p}^{2}=361$

$p=\pm \sqrt{361}$

$\therefore \mathrm{p}=\pm 19$

hence concluded that the value of p=$\pm 19$

Answer 2

Given: α,β are the zeroes of the quadratic equation $f(x)=x^2+px+48$

            $(\alpha -\beta )^2=169$

To find: The value of $p$

Solution:

To calculate the value of $p$ we need to find out the sum and the product of the roots or zeroes of the equation.

The sum of the roots is given by,

Sum of roots $(\alpha +\beta )$ = coefficient of the $x$ term/coefficient of $x^2$ = $p$

The product of the roots is given by,

Product of roots $(\alpha \beta )$ = constant term of the equation/coefficient of $x^2$ = $48$

It is given $(\alpha -\beta )^2=169$ , the $(\alpha -\beta )^2$ can be written as

⇒ $(\alpha +\beta )^2-4\alpha \beta =169$

⇒ $(-p)^2-4*48=169$

⇒ $p^2-192=169$

⇒ $p^2=169+192$

⇒ $p^2=361$

⇒$p=\sqrt{361}$

⇒ $p = +19,-19$

Final answer:

The value of p is +19 and -19.