How to solve $(\sqrt{x}+7)(\sqrt{x}-1)=\frac{105}{4}$?

Question

I'm having difficulties solving $(\sqrt{x}+7)(\sqrt{x}-1)=\frac{105}{4}$. I guess I'm missing a crucial point. The solution is $x=\frac{49}{4}$. What are the steps?

Answer 1

Let $t = \sqrt x$.

Then we can convert the question to

$(t + 7)(t - 1) = \frac{105}{4}$

Simplify it to get the following quadratic in $t$

$4t^2 + 24t - 133 = 0$

There are many ways to solve a quadratic.

By any method, you will get the following solutions for $t$

$t_1 = \frac{7}{2}$

$t_2 = \frac{-19}{2}$

From here, we can solve for $x $ to get the desired answer. Note that we want $t$ to be positive therefore we do not consider $t_2$.

Answer 2

As the user Dark Magician said, we can use the substitution $t=\sqrt{x}$. We get the following from using the quadratic formula:

\begin{align} (t+7)(t-1)&=\frac{105}{4}\\ t^2-t+7t-7&=\frac{105}{4}\\ t^2+6t-7-\frac{105}{4}&=0\\ t^2+6t-\frac{133}{4}&=0\\ 4t^2+24t-133&=0\\ t&=\frac{-24\pm\sqrt{24^2-4\times4\times(-133)}}{2\times4}\\ t&=\frac{-24\pm\sqrt{576+2128}}{8}\\ t&=\frac{-24\pm\sqrt{2704}}{8}\\ t&=-3\pm\frac{13}{2}\\ t&=-\frac{19}{2}\ or\ \frac{7}{2} \end{align}

We know that $t$ is nonnegative, as it is an arithmetic square root, so

$$t=\frac{7}{2}$$ $$x=t^2=\frac{49}{4}$$

Answer 3

The question setter has the intention of requiring a "$t = \sqrt x + 3$" substitution such that the original becomes $(t + 4) (t - 4) = 105/14$.

Then, $t^2 - 4^2 = 105/4$

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