4x-2 = x^2-3x-8

Understand the Problem

This question requires solving a quadratic equation. The goal is to find the values of 'x' that satisfy the equation 4x - 2 = x^2 - 3x - 8. This involves rearranging the equation into the standard quadratic form (ax^2 + bx + c = 0) and then solving for x, which can be done by factoring, completing the square, or using the quadratic formula.

Answer

$x = \frac{7 + \sqrt{73}}{2}$ and $x = \frac{7 - \sqrt{73}}{2}$

Steps to Solve

Rearrange the equation

Move all terms to one side of the equation to set it equal to zero. This puts the quadratic equation in standard form ($ax^2 + bx + c = 0$).

$4x - 2 = x^2 - 3x - 8$

Subtract $4x$ from both sides: $-2 = x^2 - 7x - 8$

Add $2$ to both sides: $0 = x^2 - 7x - 6$

Apply the quadratic formula

Since the quadratic equation $x^2 - 7x - 6 = 0$ does not factor easily, we use the quadratic formula to find the solutions for $x$. The quadratic formula is:

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

In our equation, $a = 1$, $b = -7$, and $c = -6$. Plugging these values into the quadratic formula gives:

$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(-6)}}{2(1)}$

Simplify the expression

Simplify the expression inside the square root and the rest of the equation:

$x = \frac{7 \pm \sqrt{49 + 24}}{2}$

$x = \frac{7 \pm \sqrt{73}}{2}$

State the solutions

The two solutions for $x$ are:

$x = \frac{7 + \sqrt{73}}{2}$ and $x = \frac{7 - \sqrt{73}}{2}$

More Information

The solutions are irrational numbers because the discriminant ($b^2 - 4ac = 73$) is not a perfect square.

Tips

Incorrectly identifying the coefficients a, b, and c.
Making errors when substituting values into the quadratic formula.
Arithmetic errors when simplifying the expression, especially under the square root.
Forgetting the $\pm$ sign, which leads to only one solution instead of two.

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