To determine which values are solutions to the quadratic equation [tex]\( x^2 + 2x = 8 \)[/tex], we first need to rewrite the equation in its standard form, which is [tex]\( ax^2 + bx + c = 0 \)[/tex].
Starting with the given equation:
[tex]\[ x^2 + 2x = 8 \][/tex]
Subtract 8 from both sides to set the equation to 0:
[tex]\[ x^2 + 2x - 8 = 0 \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -8 \)[/tex].
Next, we solve this quadratic equation using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plugging in [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -8 \)[/tex]:
[tex]\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{4 + 32}}{2} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{36}}{2} \][/tex]
[tex]\[ x = \frac{-2 \pm 6}{2} \][/tex]
We now compute the two possible solutions:
[tex]\[ x_1 = \frac{-2 + 6}{2} = \frac{4}{2} = 2 \][/tex]
[tex]\[ x_2 = \frac{-2 - 6}{2} = \frac{-8}{2} = -4 \][/tex]
Thus, the solutions to the quadratic equation [tex]\( x^2 + 2x - 8 = 0 \)[/tex] are [tex]\( x = 2 \)[/tex] and [tex]\( x = -4 \)[/tex].
Now we check which of the given options match these solutions:
A. 6
B. -1
C. -4
D. 2
E. 8
From our calculation, the solutions are 2 and -4. Therefore, the correct answers are:
C. -4
D. 2
