Which of the following equations has a solution in common with x^2 - 2x - 15 = 0?
A. x^2 - 6x + 9 = 0
B. x^2 + 2x - 15 = 0
C. 2x^2 + 9x + 9 = 0
D. 2x^2 + x - 3 = 0
E. none of the above
The OA is C.
Is there a fast way to solve this PS question? I think it could take me more than 7 minutes to solve it. <i class="em em-frowning"></i>
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Which of the following equations has a solution in common
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Vincen
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Here, we have to find the solutions of the given equation, that is to say $$x^2-2x-15=0\ \ \ \Rightarrow\ \ \left(x-5\right)\left(x+3\right)=0.$$ Therefore, the solutions are x=5 and x=-3. Now, the given options can be factorized as follows:VJesus12 wrote: Which of the following equations has a solution in common with x^2 - 2x - 15 = 0?
A. x^2 - 6x + 9 = 0
B. x^2 + 2x - 15 = 0
C. 2x^2 + 9x + 9 = 0
D. 2x^2 + x - 3 = 0
E. none of the above
The OA is C.
Is there a fast way to solve this PS question? I think it could take me more than 7 minutes to solve it. <i class="em em-frowning"></i>
$$\left(A\right)\ \ x^2-6x+9=0\ \Rightarrow\ \ \ \left(x-3\right)^2=0\ \Rightarrow\ \ x=3.$$ Incorrect. $$\left(B\right)\ \ x^2+2x-15=0\Rightarrow\ \ \ \left(x-3\right)\left(x+5\right)=0\ \Rightarrow\ \ x=3\ or\ \ x=-5.$$ Incorrect.
$$\left(C\right)\ \ 2x^2+9x+9=0\ \Rightarrow\ x_{1,2}=\frac{-9\pm\sqrt{81-72}}{4}=\frac{-9\pm3}{4}$$ $$x_1=\frac{-9+3}{4}=\frac{-6}{4}=-\frac{3}{2}.$$ $$x_2=\frac{-9-3}{4}=\frac{-12}{4}=-3.$$ Correct.
Therefore, the answer is C.
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Brent@GMATPrepNow
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Step 1: Solve the given equation: x² - 2x - 15 = 0VJesus12 wrote:Which of the following equations has a solution in common with x² - 2x - 15 = 0?
A. x² - 6x + 9 = 0
B. x² + 2x - 15 = 0
C. 2x² + 9x + 9 = 0
D. 2x² + x - 3 = 0
E. none of the above
This is a quadratic set equal to zero, so let's factor to get: (x - 5)(x + 3)=0
So, we have two solutions (roots): x = 5 or x = -3
Step 2: Solve the equations in the answer choices to see which one has a root (solution) of x = 5 or x = -3
A. x² - 6x + 9 = 0
Factor to get: (x - 3)(x - 3) = 0
So, x = 3
No shared solutions (roots) so keep moving.
B. x² + 2x - 15 = 0
Factor: (x + 5) (x - 3) = 0
So, x = -5 or x = 3
No shared solutions (roots) so keep moving.
C. 2x² + 9x + 9 = 0
This one isn't as easy to factor. So, we can try to factor it OR plug in x = 5 to x = -3 see if either is a solution to 2x² + 9x + 9 = 0
Plug in x = 5 to get 2(5)² + 9(5) + 9 = 0
Evaluate to get: 50 + 45 + 9 = 0. Doesn't work
Plug in x = -3 to get 2(-3)² + 9(-3) + 9 = 0
Evaluate to get: 18 - 27 + 9 = 0 WORKS!!
Since x = -3 is ALSO a solution to answer choice C, the correct answer is C
Cheers,
Brent
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Jeff@TargetTestPrep
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Factoring the given equation, we have:VJesus12 wrote:Which of the following equations has a solution in common with x^2 - 2x - 15 = 0?
A. x^2 - 6x + 9 = 0
B. x^2 + 2x - 15 = 0
C. 2x^2 + 9x + 9 = 0
D. 2x^2 + x - 3 = 0
E. none of the above
x^2 - 2x - 15 = 0
(x - 5)(x + 3) = 0
x = 5 or x = -3
Let's factor down each answer choice:
A. x^2 - 6x + 9 = 0
(x - 3)(x - 3) = 0
x = 3
Answer choice A is not correct.
B. x^2 + 2x - 15 = 0
(x + 5)(x - 3) = 0
x = -5 or x = 3
Answer choice B is not correct.
C. 2x^2 + 9x + 9 = 0
(2x + 3)(x + 3) = 0
x = -3/2 or x = -3
We see that this equation has a solution in common with x^2 - 2x - 15 = 0, and thus answer choice C is correct.
Answer: C
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deloitte247
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solution to $$x^2-2x-15=0$$
$$x^2+3x-5x-15=0$$
$$\left(x^2+3x\right)-\left(5x+15\right)$$
$$x\left(x+3\right)-5\left(x+3\right)$$
$$\left(x-5\right)\left(x+3\right)=0$$
$$x=5\ or\ -3$$
For two quadratic equations to have the same solutions;
(i) we must be able to express one as the factor of the other
$$e.g\ x^2-4=0\ and\ 2x^2-8=0$$
$$we\ can\ \exp ress\ 2x^2-8\ as\ 2\left(x^2-4\right)\ which\ is\ a\ factor\ of\ the\ first\ equation.$$
None of the options given can be expressed as a rational or irrational factor of the given equation.
Therefore, Option E (none of the above is the correct answer)
Note: Option B has solutions (-5 and 3) which are different from (5 and -3)
$$x^2+3x-5x-15=0$$
$$\left(x^2+3x\right)-\left(5x+15\right)$$
$$x\left(x+3\right)-5\left(x+3\right)$$
$$\left(x-5\right)\left(x+3\right)=0$$
$$x=5\ or\ -3$$
For two quadratic equations to have the same solutions;
(i) we must be able to express one as the factor of the other
$$e.g\ x^2-4=0\ and\ 2x^2-8=0$$
$$we\ can\ \exp ress\ 2x^2-8\ as\ 2\left(x^2-4\right)\ which\ is\ a\ factor\ of\ the\ first\ equation.$$
None of the options given can be expressed as a rational or irrational factor of the given equation.
Therefore, Option E (none of the above is the correct answer)
Note: Option B has solutions (-5 and 3) which are different from (5 and -3)
