Question

1. $5x^2 - 5 = 0$

$a = $ $b = $ $c = $
substitute into the quadratic formula:
what are the x - intercepts? ,

Explanation:

Step1: Identify coefficients

The quadratic equation is \(5x^2 - 5 = 0\), which is in the form \(ax^2 + bx + c = 0\). So, \(a = 5\), \(b = 0\) (since there's no \(x\) term), \(c = -5\).

Step2: Apply quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). Substituting \(a = 5\), \(b = 0\), \(c = -5\):
\(x=\frac{-0\pm\sqrt{0^2 - 4\times5\times(-5)}}{2\times5}=\frac{\pm\sqrt{100}}{10}=\frac{\pm10}{10}\)

Step3: Solve for x

For the plus sign: \(x=\frac{10}{10}=1\)
For the minus sign: \(x=\frac{-10}{10}=-1\)

Answer:

a = 5, b = 0, c = -5; x-intercepts are -1, 1