Question

which shows the correct substitution of the values a, b, and c from the equation -2 = -x + x² - 4 into the quadratic formula?
quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
\\(\circ\\) $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-4)}}{2(1)}$
\\(\circ\\) $x = \frac{-1 \pm \sqrt{1^2 - 4(-1)(-4)}}{2(-1)}$
\\(\circ\\) $x = \frac{-1 \pm \sqrt{(1)^2 - 4(-1)(-2)}}{2(-1)}$
\\(\circ\\) $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)}$

Explanation:

Step1: Rewrite the equation in standard form

The standard form of a quadratic equation is \(ax^{2}+bx + c = 0\). Given the equation \(-2=-x + x^{2}-4\), we can rewrite it by moving all terms to one side. Add \(2\) to both sides and rearrange the terms:
\(x^{2}-x - 4 + 2=0\), which simplifies to \(x^{2}-x - 2 = 0\).
Here, \(a = 1\) (coefficient of \(x^{2}\)), \(b=- 1\) (coefficient of \(x\)), and \(c=-2\) (constant term).

Step2: Substitute into the quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
Substituting \(a = 1\), \(b=-1\), and \(c = - 2\) into the formula:

• For \(-b\), we have \(-(-1)\) (since \(b=-1\)).
• For \(b^{2}\), we have \((-1)^{2}\) (since \(b = - 1\)).
• For \(4ac\), we have \(4(1)(-2)\) (since \(a = 1\) and \(c=-2\)).
• For \(2a\), we have \(2(1)\) (since \(a = 1\)).

So the substitution gives \(x=\frac{-(-1)\pm\sqrt{(-1)^{2}-4(1)(-2)}}{2(1)}\).

Answer:

\(x=\frac{-(-1)\pm\sqrt{(-1)^{2}-4(1)(-2)}}{2(1)}\) (the last option)