broall wrote:
EgmatQuantExpert wrote:
What is the product of all values of x that satisfies the equation: \(\sqrt{5x} + 1= \sqrt{(7x - 3)}\) ?
A. 9
B. 11
C. 49
D. 99
E. None of the above
First, make sure that \(\sqrt{5x}\) and \(\sqrt{7x-3}\) have real value so \(x \geq 0\) and \(x \geq \frac{3}{7}\).
Hence \(x \geq \frac{3}{7}\).Now, solve that equation
\(\begin{align}
\quad \sqrt{5x}+1 &= \sqrt{7x-3} \\
5x + 2\sqrt{5x}+1 &= 7x-3 \\
2\sqrt{5x} &= 2x-4 \\
\sqrt{5x} &= x-2 \\
x - \sqrt{5x} - 2 &=0
\end{align}\)
Set \(t = \sqrt{x} \geq \sqrt{3/7}\) we have \(t^2 -t\sqrt{5} -2 =0\)
\(\delta = 5 + 4*2 = 13 \implies t_{12}=\frac{\sqrt{5} \pm \sqrt{13}}{2}\)
We have \(t_1 = \frac{\sqrt{5}+ \sqrt{13}}{2} \approx 2.9 > 1 > \sqrt{\frac{3}{7}}\). Choose this root.
\(t_2 = \frac{\sqrt{5}- \sqrt{13}}{2} \approx -0.7 < 0 < \sqrt{\frac{3}{7}}\). Eliminate this root.
Hence \(x=t_1^2=\frac{9+\sqrt{65}}{2}\).
The answer E.
It's actually
x=(9+√65)/2
and
x=(9-√65)/2
There are 2 solutions
the best way to check it is b^2-4ac
4 possible scenarios, one of them if the result is positive non-perfect square number then there are 2 irrational solutions but the product of them will be rational number, in that case 4
(9+√65)/2*(9-√65)/2=(81-65)/4=4
the question wasn't formulated right
I've spent like 3 or 4 minutes double and cross checking myself, 'cause i got the solution here it's but not among the available answers...