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

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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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Last edited by Bunuel on 19 Sep 2017, 01:45, edited 1 time in total.
Edited the OA.

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New post 30 Jul 2017, 13:42
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Reserving this space to post the official solution. :)
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Re: What is the product of all values of x that satisfies the equation: [#permalink]

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New post 31 Jul 2017, 01:51
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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.
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Re: What is the product of all values of x that satisfies the equation: [#permalink]

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New post 31 Jul 2017, 11:38
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The final equation i am getting is x^2-9x+4
This means no solution. EgmatQuantExpert please shed some light on the solution.

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Re: What is the product of all values of x that satisfies the equation: [#permalink]

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New post 01 Aug 2017, 23:33
i am getting x^2-9x+4=0 which has no specific solution so answer should be E

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Re: What is the product of all values of x that satisfies the equation: [#permalink]

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New post 02 Aug 2017, 05:19
x^2-9x+4 is the final equation I get.
Is the answer E?

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Re: What is the product of all values of x that satisfies the equation: [#permalink]

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New post 02 Aug 2017, 10:43
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

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Re: What is the product of all values of x that satisfies the equation: [#permalink]

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New post 02 Aug 2017, 11:20
VachePBH wrote:
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


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

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Re: What is the product of all values of x that satisfies the equation: [#permalink]

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New post 02 Aug 2017, 11:37
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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.

Did in the same way, except for the last part. t~2,9, x=t^2, therefore x~9. Just one root -> ans. is A

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Re: What is the product of all values of x that satisfies the equation: [#permalink]

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New post 03 Aug 2017, 12:28
I am getting E as my final answer. After squaring both sides, I get x^2-9x+4. The product is 4 which is E in our case. I am confused as to how A is the correct answer? Could anyone please explain it to me?

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Re: What is the product of all values of x that satisfies the equation: [#permalink]

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New post 17 Aug 2017, 21:48
EgmatQuantExpert wrote:
Reserving this space to post the official solution. :)



Hi eGMAT,

Can you please elaborate on the answer?

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New post 19 Aug 2017, 04:42
As everyone solved, I am getting final equation as \(x^2-9x+4=0\)
This gives product of roots as 4 thus E is answer as further solving above equation gives roots which are slighly more than 3/7.

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Re: What is the product of all values of x that satisfies the equation: [#permalink]

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New post 19 Aug 2017, 08:58
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Probably it must be E , not A.
It will be A only if the question asked about sum of roots.

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Re: What is the product of all values of x that satisfies the equation: [#permalink]

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New post 20 Aug 2017, 01:08
The Answer provided is wrong!
it should be E and not A

EgmatQuantExpert - Kindly Rectify this

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What is the product of all values of x that satisfies the equation: [#permalink]

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New post 22 Sep 2017, 20:32
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...

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What is the product of all values of x that satisfies the equation:   [#permalink] 22 Sep 2017, 20:32
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