Answer to Question #346439 in Algebra for jalynn

Question

Are there any values of p such that p²+48 is equal to -14p?

Answer 1

If p²+48 is equal to -14p then we have the next equation:

$p^2+48=-14p,$

$p^2+14p+48=0.$

To solve this quadratic equation, let's find its discrininant:

$D=b^2-4ac=14^2-4\cdot 1\cdot 48=196-192=4.$

As $D>0$ the equation has two real distinct roots:

$p_1=\frac{-b-\sqrt{D}}{2a}=\frac{-14-2}{2}=-8,$

$p_2=\frac{-b+\sqrt{D}}{2a}=\frac{-14+2}{2}=-6.$

So, there are two values of p: -8 and -6, such that p²+48 is equal to -14p:

At $p=-8:$

$(-8)^2+48=64+48=112=-14\cdot (-8).$

At $p=-6:$

$(-6)^2+48=36+48=84=-14\cdot (-6).$

Answer: p=-8 and p=-6.