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

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

• Please use MathJax. Here is a tutorial. Dec 10, 2023 at 17:08
• Please write better. It is even hard to edit. Dec 10, 2023 at 17:10
• For $x=-3$ I obtain $\sqrt{0}+\sqrt{-20}=\sqrt{-20}$. This looks fine, so the book is right. Dec 10, 2023 at 17:14
• $1$ is definitely a solution. Plugging in $x=-3$, the equation becomes $\sqrt{-20}=\sqrt{-20}$. Whether this is true depends on how you interpret square roots of negative numbers.
– Karl
Dec 10, 2023 at 17:16
• @codeing_monkey Actually, it is the book which says that it is right. So it is not our responsibility whether or not we are taking too much liberty. One could have both opinions. Dec 10, 2023 at 17:59

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.

• I don't think so. This is only convention, which isn't mentioned in the question. The domain for $x$ is real numbers. Nevertheless $\sqrt{-n}=\sqrt{-n}$ for $n\in \Bbb N$ is certainly true. We may forbid to write that expression, but why should we? Even if we don't know what it is, it is a tautology. Dec 10, 2023 at 17:56
• @IbrarTahir -3 is a complex root while 1 is a real root. Both are roots of the equation. Like Dietrich implies, whether or not you allow for complex roots is entirely dependent on the context of the question. Dec 10, 2023 at 18:00
• @IbrarTahir I am assuming you are working in the domain of real numbers. Therefore, only $x = 1$ is a root. If you extend your domain to complex numbers, then $x = -3$ is also a root because, in this field, square roots of negative numbers are well-defined. Dec 10, 2023 at 18:16
• I would have to say then you are probably working in the reals since, by definition, the domain of the square root function is $[0, +\infty)$. Dec 10, 2023 at 18:30
• @IbrarTahir Please consider accepting my answer if you think it helped you! Dec 10, 2023 at 19:15

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 $$x = 1$$.

• Both are roots? Or only x=1 root Dec 10, 2023 at 18:11
• @IbrarTahir $x = 1$ is the unique root of the equation.
– NN2
Dec 10, 2023 at 18:21

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?

• I solve it my question is that x=1 and x=-3 both are roots or only x=1 is root Dec 10, 2023 at 18:10
• Whether you can cancel $\sqrt{x-1}$ without the assumption $x>1$ seems to be at the core of the OP's question. Dec 10, 2023 at 18:35