What is the product of all values of x that satisfies the equation:

Question

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

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niks18 · Accepted Answer

Hi  CAMANISHPARMAR

I think you are trying to calculate the roots of the equation, which is actually not required here.

you simply need to square both sides of the equation twice to get  x^2-9x+4

Now Product of roots  = \frac{c}{a} , hence  =\frac{4}{1}=4

This concept is frequently tested in GMAT and this question is actually very simple and can be solved under 2 minutes.

broall · Answer

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.

VachePBH · Answer

Why don't you calculate the Discriminant.

The answer is obvious A (9)

As most of you got the final equation, x^2-9x+4=0, this needs further modification.

In equations ax^2 + bx + c = 0, the roots can be found with Discriminant.

D = b^2 - 4ac
 
then, x1 = ((-b) - sqrt(D))/2
x2 = ((-b) + sqrt(D))/2

So in our example

D = 81 - 16 = 65

X1 = (9-sqrt(65))/2
X2 = (9+sqrt(65))/2

Summing these two will eliminate the sqrt parts, and we will have 18/2 = 9

quantumliner · Answer

What you have calculated appears to be the sum of the roots. Question asks for the Product of the roots.

cbh · Answer

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...

CAMANISHPARMAR · Answer

The answer is E only but I am not happy with the quality of this question. It is very not a GMAT like question.

Experts ... kindly clarify!!

pepperoniPizza · Answer

Since both sides are non-negative we can square them:
We then have 5x+1+2 \sqrt{5x}  = 7x -3
Or x -  \sqrt{5x}  - 2 = 0

We know that for quadratic equation a x^2  + bx +c = 0, if Delta =  b^2  - 4ac > 0 the equation will have 2 roots, hence the equation above has 2 roots.
According to Viéte formula, the product of the roots equals  \frac{c}{a}  or -2 in this case => Hence E

circa123 · Answer

Given, √5x+1=√(7x−3) 
Squaring Both sides we get
5x+1+2√5x = 7x -3
Or x - √5x - 2 = 0

For a quadratic equation ax^2 + bx +c = 0, we know that 
sum of roots is equal to b and product of roots is c/a.

Subsequently we can say that product of roots of √x is -2. for x is 4.

Option E is the correct answer.

Deepa444 · Answer

chetan2u  can u help with the solution?

s3sanghvi · Answer

I solved in the below manner,

= \sqrt{5x} + 1 = \sqrt{7x-3}

squaring both sides,

=> ( \sqrt{5x} +1)^2 = 7x-3
=> 5x + 1 + 2\sqrt{5x} = 7x-3 
=> 2\sqrt{5x} = 2x-4
=> \sqrt{5x} = x-2

Now we can replace the

\sqrt{5x} in the above initial equation as x-2

=> (x-2+1) = \sqrt{7x-3}
=> x-1= \sqrt{7x-3}

squaring both sides,

=> (x-1)^2 = 7x-3
=> x^2+1-2x = 7x-3
=> x^2-9x+4=0

Now product of the solution of a quadratic equation is c/a 
=> 4/1 => 4

Hence E is the answer.

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What is the product of all values of x that satisfies the equation:

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