Question

1. $6x^2 - 18x - 18 = 6$
2. $x^2 + 5x - 35 = 3x$
3. $3x^2 - 16x - 7 = 5$
4. $7x^2 + 2x = 0$

Explanation:

Problem 9: \(6x^2 - 18x - 18 = 6\)

Step 1: Simplify the equation

Subtract 6 from both sides: \(6x^2 - 18x - 24 = 0\)
Divide all terms by 6: \(x^2 - 3x - 4 = 0\)

Step 2: Factor the quadratic

Find two numbers that multiply to -4 and add to -3: -4 and 1.
So, \((x - 4)(x + 1) = 0\)

Step 3: Solve for \(x\)

Set each factor to zero:
\(x - 4 = 0 \implies x = 4\)
\(x + 1 = 0 \implies x = -1\)

Problem 12: \(x^2 + 5x - 35 = 3x\)

Step 1: Simplify the equation

Subtract \(3x\) from both sides: \(x^2 + 2x - 35 = 0\)

Step 2: Factor the quadratic

Find two numbers that multiply to -35 and add to 2: 7 and -5.
So, \((x + 7)(x - 5) = 0\)

Step 3: Solve for \(x\)

Set each factor to zero:
\(x + 7 = 0 \implies x = -7\)
\(x - 5 = 0 \implies x = 5\)

Problem 15: \(3x^2 - 16x - 7 = 5\)

Step 1: Simplify the equation

Subtract 5 from both sides: \(3x^2 - 16x - 12 = 0\)

Step 2: Factor the quadratic (or use quadratic formula)

Using quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 3\), \(b = -16\), \(c = -12\)
Discriminant: \(b^2 - 4ac = (-16)^2 - 4(3)(-12) = 256 + 144 = 400\)
\(\sqrt{400} = 20\)
\(x = \frac{16 \pm 20}{6}\)

Step 3: Solve for \(x\)

First solution: \(x = \frac{16 + 20}{6} = \frac{36}{6} = 6\)
Second solution: \(x = \frac{16 - 20}{6} = \frac{-4}{6} = -\frac{2}{3}\)

Problem 18: \(7x^2 + 2x = 0\)

Answer:

s:

• Problem 9: \(x = 4\) or \(x = -1\)
• Problem 12: \(x = -7\) or \(x = 5\)
• Problem 15: \(x = 6\) or \(x = -\frac{2}{3}\)
• Problem 18: \(x = 0\) or \(x = -\frac{2}{7}\)