Finding roots of the equation $\sqrt{x^2+2x-3}+\sqrt{x^2+7x-8}=\sqrt{5(x^2+3x-4)}$

Question

Many helping books and also book mention the root of equation is $1$ and $-3$, but I think there is only root $1$. $$\sqrt{x^2+2x-3}+\sqrt{x^2+7x-8}=\sqrt{5(x^2+3x-4)}$$

Answer 1

When you solve the equation by removing square roots you obtain the equation $$4x^4 + 36x^3 + 12x^2 - 148x + 96 = 9x^4 + 36x^3 - 18x^2 - 108x + 81$$ which gives you $x = 1, -3$. However, we must verify that these are indeed solutions to our original equation (with square roots) and you are correct that only $x = 1$ is true. Of course, this is assuming you are working only in $\mathbb{R}$ and not $\mathbb{C}$ since you have indicated that this is a question for a book on Pre-Calculus.

Answer 2

HINT

Cancel common factor $(x-1)$ from under the radical $x\ne 1$.Then $$\sqrt{x+3}+\sqrt{x+8}=\sqrt{5(x+4)}$$

Can you square and simplify?

Answer 3

Denote $a = \sqrt{x^2+2x-3}$ and $b = \sqrt{x^2+7x-8}$, we have: $$\begin{align} &5(x^2+3x-4) = 4(x^2+2x-3) +(x^2+7x-8) = 4a^2+b^2\\ &\iff a+b = \sqrt{4a^2+b^2}\\ &\iff a^2+2ab+b^2 =4a^2+b^2\\ &\iff \cases{a=0 \\2b = 3a}\\ &\iff \cases{(1): x^2+2x-3 = 0 \\(2):4(x^2+7x-8) = 9(x^2+2x-3)} \end{align}$$

The case $(1)$ has two roots $x = 1$ and $x = -3$.

The case $(2)$ has 1 root $x = 1$.

We need to put $x = 1$ and $x =-3$ to test the equation. Only the root $x = 1$ satisfies the equation (the roots $x = -3$ make the second term $x^2+7x-8$ negative, so we need to remove it).

By consequence, only the equation has only $1$ root $x = 1$.