Question

1. $-x^2 + 25$
2. $8x^2 - 98$
3. $x^2 - 1$
4. $-64x^2 + 121$
5. $28x^2 - 7$
6. $27x^2 - 12$

d) solve each quadratic equation.

1. $(x - 6)(2x + 1) = 0$
2. $x^2 - x - 72 = 0$
3. $x^2 - 16 = 0$
4. $x^2 = 6x$
5. $x(x - 12) = -36$
6. $(x + 14)(x - 3) = 0$

Explanation:

Problem 34: \((x - 6)(2x + 1) = 0\)

Step 1: Apply Zero - Product Property

The zero - product property states that if \(ab = 0\), then either \(a = 0\) or \(b = 0\) (or both). For the equation \((x - 6)(2x+1)=0\), we set each factor equal to zero:
\(x - 6 = 0\) or \(2x + 1=0\)

Step 2: Solve for \(x\) in each equation

• For \(x - 6=0\), add 6 to both sides of the equation: \(x=6\)
• For \(2x + 1 = 0\), subtract 1 from both sides: \(2x=-1\), then divide both sides by 2: \(x =-\frac{1}{2}\)

Step 1: Factor the quadratic expression

We need to find two numbers that multiply to \(- 72\) and add up to \(-1\). The numbers are \(-9\) and \(8\) because \((-9)\times8=-72\) and \(-9 + 8=-1\). So we can factor the quadratic as:
\(x^{2}-x - 72=(x - 9)(x + 8)=0\)

Step 2: Apply Zero - Product Property

Set each factor equal to zero: \(x - 9 = 0\) or \(x + 8=0\)

Step 3: Solve for \(x\)

• For \(x - 9=0\), we get \(x = 9\)
• For \(x + 8=0\), we get \(x=-8\)

Step 1: Rewrite the equation as a difference of squares

We know that \(a^{2}-b^{2}=(a - b)(a + b)\). Here, \(x^{2}-16=x^{2}-4^{2}\), so we can factor it as \((x - 4)(x + 4)=0\)

Step 2: Apply Zero - Product Property

Set each factor equal to zero: \(x - 4=0\) or \(x + 4=0\)

Step 3: Solve for \(x\)

• For \(x - 4=0\), we get \(x = 4\)
• For \(x + 4=0\), we get \(x=-4\)

Answer:

\(x = 6\) or \(x=-\frac{1}{2}\)

Problem 35: \(x^{2}-x - 72=0\)