Question

1. $x^2 - 4x = -4$

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

Explanation:

First, we need to rewrite the equation \(x^{2}-4x = - 4\) in the standard quadratic form \(ax^{2}+bx + c = 0\).

Step 1: Rewrite the equation

We add 4 to both sides of the equation \(x^{2}-4x=-4\) to get \(x^{2}-4x + 4=0\).

In the standard quadratic form \(ax^{2}+bx + c = 0\), we can identify the coefficients:

• For the coefficient of \(x^{2}\), we have \(a = 1\) (since the coefficient of \(x^{2}\) in \(x^{2}-4x + 4\) is 1).
• For the coefficient of \(x\), we have \(b=- 4\) (the coefficient of \(x\) is - 4).
• For the constant term, we have \(c = 4\) (the constant term is 4).

Now, we use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) to find the roots (which are the x - intercepts).

Step 2: Substitute the values of \(a\), \(b\), and \(c\) into the quadratic formula

Substitute \(a = 1\), \(b=-4\) and \(c = 4\) into the quadratic formula:

\(x=\frac{-(-4)\pm\sqrt{(-4)^{2}-4\times1\times4}}{2\times1}\)

Step 3: Simplify the expression

First, simplify the numerator:

• \(-(-4)=4\)
• \((-4)^{2}-4\times1\times4=16 - 16=0\)

So, \(x=\frac{4\pm\sqrt{0}}{2}=\frac{4\pm0}{2}\)

Step 4: Find the value of \(x\)

When we take the plus - minus sign, we have two cases:

• Case 1: \(x=\frac{4 + 0}{2}=\frac{4}{2}=2\)
• Case 2: \(x=\frac{4-0}{2}=\frac{4}{2}=2\)

Since both solutions are the same, the quadratic equation has a repeated root at \(x = 2\).

Answer:

The value of \(a = 1\), \(b=-4\), \(c = 4\). The x - intercept is \(x = 2\) (with multiplicity 2).