# The set of all $x$ satisfying, $\sqrt{4x+1} + \sqrt{7-x} = 6$, consists of:

The set of all $$x$$ satisfying, $$\sqrt{4x+1} + \sqrt{7-x} = 6$$, consists of:

$$A)$$ Two rational numbers. $$B)$$ An irrational number. $$C)$$ Complex number. $$D)$$ None.

How to solve the above question using proper method?

I got the values by substitution. I just considered the values after evaluation of the radical to be $$(5,1)$$ ,$$(3,3)$$ and $$(4,2)$$.

This leads me to nice and whole values of $$x = 6$$. I can't find any other solutions.

I tried squaring, but the solution is getting really nasty at a point. I wonder if there could be any more methods to solve this problem. Any help or suggestion would be appreciated. Thanks!

• do you mean, $\sqrt{4x+1}$ instead of $\sqrt{4x-1}$? Feb 27, 2021 at 11:47
• @Destiny If you know trigonometry then there is a nice method to solve such equations Feb 27, 2021 at 11:48
• It has two rational roots over $[-\frac 14,7]\\$ desmos.com/calculator/lxb2nau18g Feb 27, 2021 at 11:52

## 3 Answers

Note that\begin{align}\sqrt{4x+1}+\sqrt{7-x}=6&\implies4x+1+7-x+2\sqrt{4x+1}\sqrt{7-x}=36\\&\iff2\sqrt{4x+1}\sqrt{7-x}=28-3x\\&\implies4(4x+1)(7-x)=(28-3x)^2\\&\iff25 x^2-276 x+756=0\\&\iff x=6\text{ or }x=\frac{126}{25}.\end{align}And you can easily check that $$6$$ and $$\frac{126}{25}$$ are actually roots of the original equation.

• +1 Very nice distinctions drawn between $\implies$ and $\iff$. Feb 27, 2021 at 12:09

WLOG $$\sqrt{4x+1}=6\cos^2t$$ and $$\sqrt{7-x}=6\sin^2t$$

$$\implies4x+1=36\cos^4t,7-x=36\sin^4t$$

$$\implies36\cos^4t+144\sin^4t=\cdots=29$$

$$\implies36\cos^4t+144(1-\cos^2t)^2=\cdots=29$$

$$\iff180\cos^4t-288\cos^2t+115=0$$

$$\implies\cos^2t=\dfrac56,\dfrac{23}{30}$$

$$\implies x=\dfrac{36(\cos^2t)^2-1}4=?$$

Here's is a variation of Jose's answer, which appeared as I was typing this up:

$$\sqrt{4x+1} + \sqrt{7-x} = 6 \\ \iff \sqrt{4x+1} = 6 - \sqrt{7-x}$$ $$\implies 4x+1 = 36+(7-x) - 12\sqrt{7-x}\tag{1} \\ \iff 5x-42=12\sqrt{7-x}$$ $$\implies 25x^2+1764-420x=144(7-x) \tag{2}\\ \iff 25x^2-276x+756=0 \\ \iff x=6\text{ or }\frac{126}{25}$$

The steps labelled (1) & (2) aren't invertible and might have created extraneous solutions; a quick check verifies that both $$6$$ and $$\frac{126}{25}$$ indeed satisfy the original equation.

• It is better to write $126/25$ to emphasize that it is a rational number, and not merely some real number rounded to two decimals $5.04$. Feb 27, 2021 at 12:22