Question #346439

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


1
Expert's answer
2022-06-01T07:58:03-0400

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

p2+48=14p,p^2+48=-14p,

p2+14p+48=0.p^2+14p+48=0.

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

D=b24ac=1424148=196192=4.D=b^2-4ac=14^2-4\cdot 1\cdot 48=196-192=4.

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

p1=bD2a=1422=8,p_1=\frac{-b-\sqrt{D}}{2a}=\frac{-14-2}{2}=-8,

p2=b+D2a=14+22=6.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:p=-8:

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

At p=6:p=-6:

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


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


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