Question

solve for \\(x\\) in the equation \\(2x^2 + 3x - 7 = x^2 + 5x + 39\\).

\\(\bigcirc\\) \\(x = -1 \pm \sqrt{47}\\)
\\(\bigcirc\\) \\(x = -1 \pm 2\sqrt{11}\\)
\\(\bigcirc\\) \\(x = 1 \pm \sqrt{53}\\)
\\(\bigcirc\\) \\(x = 1 \pm \sqrt{47}\\)

Explanation:

⚡ Using what you learned: factoring and solving quadratic equations

Step 1: Write the equation and move all terms to one side

$$ 2x^2 + 3x - 7 = x^2 - 5x + 39 $$

Subtract \( x^2 - 5x + 39 \) from both sides:

$$ (2x^2 - x^2) + (3x - (-5x)) + (-7 - 39) = 0 $$

$$ x^2 + 8x - 46 = 0 $$

Step 2: Apply the quadratic formula

For \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = 8 \), and \( c = -46 \):

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

$$ x = \frac{-8 \pm \sqrt{8^2 - 4(1)(-46)}}{2(1)} $$

$$ x = \frac{-8 \pm \sqrt{64 + 184}}{2} $$

$$ x = \frac{-8 \pm \sqrt{248}}{2} $$

Step 3: Simplify the radical and the fraction

Simplify \( \sqrt{248} \):

$$ \sqrt{248} = \sqrt{4 \times 62} = 2\sqrt{62} $$

Substitute back into the expression:

$$ x = \frac{-8 \pm 2\sqrt{62}}{2} $$

$$ x = -4 \pm \sqrt{62} $$

Answer:

\( x = -4 \pm \sqrt{62} \)