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

Question

I tried squaring both sides but doesn't seem like a good idea.

$$x^2+8x+7+\sqrt{(x^2+8x+7)(x^2+3x+2)}+x^2+3x+2=6x^2+19x+13$$ $$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4x^2+8x+4$$ $$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4(x+1)^2$$ Is there a better way for solving this equation?

Answer 1

After writing our equation in the form $$\sqrt{(x+1)(x+7)}+\sqrt{(x+1)(x+2)}=\sqrt{(x+1)(6x+13)}$$ we obtain the domain: $$(-\infty-7]\cup[-1,+\infty).$$ 1. $x\geq-1.$

We see that $-1$ is a root and it remains to solve here $$\sqrt{x+7}+\sqrt{x+2}=\sqrt{6x+13}$$ or $$\sqrt{(x+7)(x+2)}=2(x+1)$$ or $$3x^2-x-10=0,$$ which gives also $x=2.$

2. $x<-1$.

Thus, $x\leq-7$ and we need to solve $$\sqrt{-x-7}+\sqrt{-x-2}=\sqrt{-6x-13}$$ or $$\sqrt{(x+7)(x+2)}=-2(x+1),$$ which gives $$3x^2-x-10=0$$ again or $$(x-2)(3x+5)=0$$ and we see that this equation has no roots for $x\leq-7.$

Id est, we got the answer: $$\{-1,2\}$$

Answer 2

I would factor both sides first so that:

$$\sqrt{(x+7)(x+1)}+\sqrt{(x+2)(x+1)}=\sqrt{(6x+13)(x+1)}$$ $$\sqrt{x+1}(\sqrt{x+7}+\sqrt{x+2})=\sqrt{x+1}\sqrt{6x+13}$$ $$\sqrt{x+7}+\sqrt{x+2}=\sqrt{6x+13}$$ $$2x+9+2\sqrt{(x+2)(x+7)}=6x+13$$ $$\sqrt{(x+2)(x+7)}=2x+2$$ $$(x+2)(x+7)=4x^2+8x+4$$ $$3x^2-x-10=0$$ $$(3x+5)(x-2)=0$$

So $x=-\frac{5}{3}$ or $x=2$. However, you find that only $x=2$ works by putting the value into the equation. Do note that while this method may be easier, another solution, $x=-1$ has not been found which would be found by squaring both sides: https://www.symbolab.com/solver/step-by-step/%5Csqrt%7Bx%5E%7B2%7D%2B8x%2B7%7D%2B%5Csqrt%7Bx%5E%7B2%7D%2B3x%2B2%7D%3D%5Csqrt%7B6x%5E%7B2%7D%2B19x%2B13%7D. While this may not be a complete answer, I hope this helped!

Answer 3

Observe that $(x+1)$ divides all quadratics: the original equation is

$$\sqrt{(x+1)(x+7)} + \sqrt{(x+1)(x+2)} = \sqrt{(x+1)(6x+13)},$$

and rearranging we obtain the following:

$$(\sqrt{x+1})(\sqrt{x+7}+\sqrt{x+2}-\sqrt{6x+13})=0.$$

We are doing some trickery with allowing square roots to venture into $\mathbb C$ here, but note that it is all still correct: for the original equation to have solutions in $\mathbb R$, then either $x \geq -1$ (and all the square roots stay safely within $\mathbb R$) or $x \leq -7$ (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 $$\sqrt{6x+13}= \sqrt{x+7}+\sqrt{x+2},$$ and upon squaring both sides gives $$6x+13=2x+9+2\sqrt{(x+7)(x+2)},$$ which rearranges to $$\sqrt{(x+7)(x+2)}=2x+2.$$ Square again, solve the quadratic, and test your solutions to finish.

Answer 4

When you squared you forget to put the coefficient of 2.

You should have $2\sqrt{(x^2+8x+7)(x^2+3x+2)}=4(x+1)^2$

Or $\sqrt{(x^2+8x+7)(x^2+3x+2)}=2(x+1)^2$

....

But anyway....

Square again. But it'll help to factor components.

$(x^2+8x+7)(x^2+3x+2)=4(x+1)^4$

$(x+7)(x+1)(x+2)(x+1)=4(x+1)^4$.

If $x = -1$ we get $0=0$ which could be a solution although we fatored so late in the game we should check if that's extraneous.

Nope... way back in the beginning $x = -1$ would have given us $\sqrt 0 + \sqrt 0 =\sqrt 0$ which is fine.

SO $x =-1$ is a solution. Divide both sides by $(x+1)^2$ and we tet

$(x+7)(x+2)= 4(x+1)^2$ which is just a quadratic...

$x^2 + 9x + 14 = 4x^2 + 8x + 4$ so

$3x^2 -x -10=0$ which we can solve with the quadratic equation. (but well have to search for extraneous solutions.)

$x = \frac {1\pm \sqrt{1 + 120}}6 = \frac {1\pm 11}6= 2, -\frac 53$.

But $x =-\frac 53$ give $\sqrt {\frac {25}9 - \frac {40}3+7}= \sqrt{-\frac {32}9}$ which is not real.

So $x=2$ or $x=-1$