Solve \(\displaystyle\sqrt{{{x}^{{2}}+{8}{x}+{7}}}+\sqrt{{{x}^{{2}}+{3}{x}+{2}}}=\sqrt{{{6}{x}^{{2}}+{19}{x}+{13}}}\)

rijstmeel7d4t

rijstmeel7d4t

Answered question

2022-03-22

Solve x2+8x+7+x2+3x+2=6x2+19x+13

Answer & Explanation

Dixie Reed

Dixie Reed

Beginner2022-03-23Added 15 answers

Observe that (x+1) divides all quadratics: the original equation is
(x+1)(x+7)+(x+1)(x+2)=(x+1)(6x+13),
and rearranging we obtain the following:
(x+1)(x+7+x+26x+13)=0.
We are doing some trickery with allowing square roots to venture into C here, but note that it is all still correct: for the original equation to have solutions in R, then either x1 (and all the square roots stay safely within R) or x7 (in which case all of the square roots are of negative values, so the extra factors of i safely distribute out, assuming we use the principal square root).
The first factor yields a solution of -1. The second factor gives solutions when
6x+13=x+7+x+2,
and upon squaring both sides gives
6x+13=2x+9+2(x+7)(x+2),
which rearranges to
(x+7)(x+2)=2x+2.
Square again, solve the quadratic, and test your solutions to finish.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?