Norris Mcfarlane
03/18/2024 · Primary School
\( x = \frac { - ( - 2 ) \pm \sqrt { ( - 2 ) ^ { 2 } - 4 ( 1 ) ( - 6 ) } } { 2 ( 1 ) } \)
UpStudy ThothAI Solution
Tutor-Verified Answer
Step-by-step Solution
To solve the equation given in the form of the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
we can identify the coefficients from your expression:
- \( a = 1 \)
- \( b = -2 \)
- \( c = -6 \)
Now, let's substitute these values into the formula step by step.
1. Calculate \( b^2 \):
\[
b^2 = (-2)^2 = 4
\]
2. Calculate \( 4ac \):
\[
4ac = 4 \cdot 1 \cdot (-6) = -24
\]
3. Now, compute \( b^2 - 4ac \):
\[
b^2 - 4ac = 4 - (-24) = 4 + 24 = 28
\]
4. Now, substitute back into the quadratic formula:
\[
x = \frac{-(-2) \pm \sqrt{28}}{2 \cdot 1}
\]
5. Simplify:
\[
x = \frac{2 \pm \sqrt{28}}{2}
\]
6. Simplify \( \sqrt{28} \):
\[
\sqrt{28} = \sqrt{4 \cdot 7} = 2\sqrt{7}
\]
7. Substitute \( \sqrt{28} \) back into the equation:
\[
x = \frac{2 \pm 2\sqrt{7}}{2}
\]
8. Simplify further:
\[
x = 1 \pm \sqrt{7}
\]
Thus, the solutions for \( x \) are:
\[
x = 1 + \sqrt{7} \quad \text{and} \quad x = 1 - \sqrt{7}
\]
Quick Answer
The solutions for \( x \) are \( x = 1 + \sqrt{7} \) and \( x = 1 - \sqrt{7} \).
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