Question

1. what does absolute value mean? can an absolute value ever be negative?

set
for each of the quadratic equations, find the solutions using an efficient method. state the method
you used.

1. $x^{2}+17x + 60 = 0$
2. $x^{2}+16x + 39 = 0$
3. $x^{2}+7x - 5 = 0$
4. $3x^{2}+14x - 5 = 0$
5. $x^{2}-12x=-8$
6. $x^{2}+6x = 7$

Explanation:

Step1: Solve \(x^{2}+17x + 60=0\) by factoring

We need to find two numbers \(a\) and \(b\) such that \(a + b=17\) and \(a\times b = 60\). The numbers are \(12\) and \(5\).

$$x^{2}+17x + 60=(x + 12)(x+5)=0$$

Set each factor equal to zero:
\(x + 12=0\) gives \(x=-12\); \(x + 5=0\) gives \(x=-5\)

Step2: Solve \(x^{2}+16x + 39=0\) by factoring

Find two numbers \(m\) and \(n\) such that \(m + n=16\) and \(m\times n=39\). The numbers are \(13\) and \(3\)

$$x^{2}+16x + 39=(x + 13)(x + 3)=0$$

Set each factor equal to zero:
\(x+13 = 0\) gives \(x=-13\); \(x + 3=0\) gives \(x=-3\)

Step3: Solve \(x^{2}+7x-5=0\) by quadratic formula

For a quadratic equation \(ax^{2}+bx + c=0\) (\(a = 1\), \(b=7\), \(c=-5\)), the quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)
First, calculate the discriminant \(\Delta=b^{2}-4ac=(7)^{2}-4\times1\times(-5)=49 + 20=69\)
Then \(x=\frac{-7\pm\sqrt{69}}{2}\)

Step4: Solve \(3x^{2}+14x-5=0\) by factoring

We can rewrite the equation as \(3x^{2}+15x - x-5=0\)
Group the terms: \(3x(x + 5)-(x + 5)=0\), \((3x-1)(x + 5)=0\)
Set each factor equal to zero:
\(3x-1=0\) gives \(x=\frac{1}{3}\); \(x + 5=0\) gives \(x=-5\)

Step5: Solve \(x^{2}-12x=-8\) (rewrite as \(x^{2}-12x + 8=0\)) by quadratic formula

Here \(a = 1\), \(b=-12\), \(c = 8\)
The discriminant \(\Delta=(-12)^{2}-4\times1\times8=144-32 = 112\)
\(x=\frac{12\pm\sqrt{112}}{2}=\frac{12\pm4\sqrt{7}}{2}=6\pm2\sqrt{7}\)

Step6: Solve \(x^{2}+6x=7\) (rewrite as \(x^{2}+6x-7=0\)) by factoring

Find two numbers \(p\) and \(q\) such that \(p+q=6\) and \(p\times q=-7\). The numbers are \(7\) and \(-1\)

$$x^{2}+6x-7=(x + 7)(x-1)=0$$

Set each factor equal to zero:
\(x + 7=0\) gives \(x=-7\); \(x-1=0\) gives \(x = 1\)

Answer:

1. Solutions: \(x=-12\) and \(x=-5\), Method: Factoring
2. Solutions: \(x=-13\) and \(x=-3\), Method: Factoring
3. Solutions: \(x=\frac{-7+\sqrt{69}}{2}\) and \(x=\frac{-7-\sqrt{69}}{2}\), Method: Quadratic formula
4. Solutions: \(x=\frac{1}{3}\) and \(x=-5\), Method: Factoring
5. Solutions: \(x=6 + 2\sqrt{7}\) and \(x=6-2\sqrt{7}\), Method: Quadratic formula
6. Solutions: \(x=-7\) and \(x = 1\), Method: Factoring