Question

math 2: 8.2 progress check
one of these quadratics is factorable, but the other requires use
of the quadratic formula.
decide the best method, then solve each.
$x^2 + 6x - 7 = 0$ $x^2 - 4x - 7 = 0$
factorable:
quadratic form

Explanation:

For \( x^2 + 6x - 7 = 0 \) (Factorable)

Step1: Factor the quadratic

We need two numbers that multiply to \(-7\) and add to \(6\). The numbers are \(7\) and \(-1\) since \(7\times(-1)=-7\) and \(7 + (-1)=6\). So, we can factor the quadratic as:
\(x^2 + 6x - 7=(x + 7)(x - 1)=0\)

Step2: Solve for \(x\)

Set each factor equal to zero:

• \(x + 7 = 0\) gives \(x=-7\)
• \(x - 1 = 0\) gives \(x = 1\)

For \( x^2 - 4x - 7 = 0 \) (Quadratic Formula)

Step1: Identify \(a\), \(b\), \(c\)

For a quadratic equation \(ax^2+bx + c = 0\), here \(a = 1\), \(b=-4\), \(c=-7\)

Step2: Apply the quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\)
Substitute \(a = 1\), \(b=-4\), \(c=-7\) into the formula:
First, calculate the discriminant \(b^2-4ac=(-4)^2-4\times1\times(-7)=16 + 28 = 44\)
Then, \(x=\frac{-(-4)\pm\sqrt{44}}{2\times1}=\frac{4\pm2\sqrt{11}}{2}=2\pm\sqrt{11}\)

Answer:

For \(x^2 + 6x - 7 = 0\), the solutions are \(x = 1\) and \(x=-7\)
For \(x^2 - 4x - 7 = 0\), the solutions are \(x = 2+\sqrt{11}\) and \(x = 2-\sqrt{11}\)