What is the sum of all unique solutions for x^2+6x+9=|x+3| ?
A. -12
B. -9
C. -7
D. -5
E. 0
The OA is B.
When I solved this problem I got 0. Why is B the correct answer? <i class="em em-confused"></i>
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What is the sum of all unique solutions for
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Vincen
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Hi.
This is the way I'd solve it:
$$x^2+6x+9=|x+3|\ $$ $$\Rightarrow\ \ x^2+6x+9=x+3\ \ or\ \ \ x^2+6x+9=-\left(x+3\right)$$ Now, solving the first equation we get: $$x^2+5x+6=0$$ $$\Rightarrow\ \ \left(x+3\right)\left(x+2\right)=0$$ $$\Rightarrow\ x_1=-3\ ,\ \ \ x_2=-2\ \ .$$ Solving the second equation we get: $$x^2+6x+9=-\left(x+3\right)$$ $$\Rightarrow\ \ x^2+7x+12=0$$ $$\Rightarrow\ \ \left(x+4\right)\left(x+3\right)=0$$ $$\Rightarrow\ \ x_3=-4\ ,\ \ x_4=-3.$$ Now, the unique solutions are x_1=-3, x_2=-2, x_3=-4.
Therefore, $$x_1+x_2+x_3=-3-2-4=-9.$$ Hence, the correct answer is the option B.
I hope it helps you.
This is the way I'd solve it:
$$x^2+6x+9=|x+3|\ $$ $$\Rightarrow\ \ x^2+6x+9=x+3\ \ or\ \ \ x^2+6x+9=-\left(x+3\right)$$ Now, solving the first equation we get: $$x^2+5x+6=0$$ $$\Rightarrow\ \ \left(x+3\right)\left(x+2\right)=0$$ $$\Rightarrow\ x_1=-3\ ,\ \ \ x_2=-2\ \ .$$ Solving the second equation we get: $$x^2+6x+9=-\left(x+3\right)$$ $$\Rightarrow\ \ x^2+7x+12=0$$ $$\Rightarrow\ \ \left(x+4\right)\left(x+3\right)=0$$ $$\Rightarrow\ \ x_3=-4\ ,\ \ x_4=-3.$$ Now, the unique solutions are x_1=-3, x_2=-2, x_3=-4.
Therefore, $$x_1+x_2+x_3=-3-2-4=-9.$$ Hence, the correct answer is the option B.
I hope it helps you.
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ErikaPrepScholar
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$$x^2+6x+9=\left|x+3\right|$$ The absolute value gives us two options: $$x^2+6x+9=x+3$$ or $$x^2+6x+9=-\left(x+3\right)$$ $$x^2+6x+9=-x-3$$ Solving each individually:
$$x^2+6x+9=x+3$$ $$x^2+5x+6=0$$ $$\left(x+2\right)\left(x+3\right)=0$$ $$x=-2\ and\ x=-3$$
and $$x^2+6x+9=-x-3$$ $$x^2+7x+12=0$$ $$\left(x+3\right)\left(x+4\right)=0$$ $$x=-3\ and\ x=-4$$
Notice that x=-3 is a solution for both equations. We are looking for total *unique* solutions, so we'll only include it once.
So our unique solutions are -2, -3, and -4. To find the sum, we'll add them together: $$-2\ +\left(-3\right)+\left(-4\right)=-2-3-4=-9$$ This gives B as the correct answer.
$$x^2+6x+9=x+3$$ $$x^2+5x+6=0$$ $$\left(x+2\right)\left(x+3\right)=0$$ $$x=-2\ and\ x=-3$$
and $$x^2+6x+9=-x-3$$ $$x^2+7x+12=0$$ $$\left(x+3\right)\left(x+4\right)=0$$ $$x=-3\ and\ x=-4$$
Notice that x=-3 is a solution for both equations. We are looking for total *unique* solutions, so we'll only include it once.
So our unique solutions are -2, -3, and -4. To find the sum, we'll add them together: $$-2\ +\left(-3\right)+\left(-4\right)=-2-3-4=-9$$ This gives B as the correct answer.
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