Question

algebra crc skills a test answer sheet name period skill 15 skill 17 skill 18 skill 19 directions: for each of the following, use the method(s) that you find to be the easiest to answer the question provided. show all work on a separate sheet of paper. be sure to check your answers (where possible) to help you to recognize if you have made a mistake. if you use a calculator, make sure that you show the correct set up so that it is clear that you know how to solve the problem without a calculator. (2x - 3)(x + 5) = 0 y² + x = 40 (x + 3)(2x - 1) = 0 x² + 5x - 14 = 0 2x² + 7 = 169 (x - 7)² = 25 5(x + 1)² - 2 = 43 x² + 18x + ——

Explanation:

Step1: Solve \((2x - 3)(x + 5)=0\)

Using zero - product property: if \(ab = 0\), then \(a = 0\) or \(b = 0\).
Set \(2x-3 = 0\) or \(x + 5=0\).
For \(2x-3=0\), add 3 to both sides: \(2x=3\), then divide by 2: \(x=\frac{3}{2}\).
For \(x + 5=0\), subtract 5 from both sides: \(x=-5\).

Step2: Solve \((x + 3)(2x-1)=0\)

Using zero - product property.
Set \(x + 3=0\) or \(2x-1 = 0\).
For \(x + 3=0\), subtract 3 from both sides: \(x=-3\).
For \(2x-1=0\), add 1 to both sides: \(2x = 1\), then divide by 2: \(x=\frac{1}{2}\).

Step3: Solve \(2x^{2}+7 = 169\)

Subtract 7 from both sides: \(2x^{2}=169 - 7=162\).
Divide both sides by 2: \(x^{2}=\frac{162}{2}=81\).
Take square roots: \(x=\pm\sqrt{81}=\pm9\).

Step4: Solve \(5(x + 1)^{2}-2 = 43\)

Add 2 to both sides: \(5(x + 1)^{2}=43 + 2=45\).
Divide both sides by 5: \((x + 1)^{2}=\frac{45}{5}=9\).
Take square roots: \(x + 1=\pm\sqrt{9}=\pm3\).
Case 1: \(x + 1=3\), subtract 1: \(x=2\).
Case 2: \(x + 1=-3\), subtract 1: \(x=-4\).

Step5: Solve \(x^{2}+x-30 = 0\)

Factor the quadratic: find two numbers that multiply to - 30 and add to 1. The numbers are 6 and - 5.
So \(x^{2}+x - 30=(x + 6)(x - 5)=0\).
Using zero - product property: \(x+6 = 0\) or \(x - 5=0\).
\(x=-6\) or \(x = 5\).

Step6: Solve \(x^{2}+5x-14 = 0\)

Factor the quadratic: find two numbers that multiply to - 14 and add to 5. The numbers are 7 and - 2.
So \(x^{2}+5x - 14=(x + 7)(x - 2)=0\).
Using zero - product property: \(x+7 = 0\) or \(x - 2=0\).
\(x=-7\) or \(x = 2\).

Step7: Solve \((x - 7)^{2}=25\)

Take square roots: \(x - 7=\pm\sqrt{25}=\pm5\).
Case 1: \(x - 7=5\), add 7: \(x=12\).
Case 2: \(x - 7=-5\), add 7: \(x=2\).

Step8: Solve \(x^{2}+18x+\underline{}\) (complete the square)

For a quadratic \(x^{2}+bx\), to complete the square, we add \((\frac{b}{2})^{2}\). Here \(b = 18\), so \((\frac{18}{2})^{2}=81\). So the missing term is 81, and the quadratic becomes \((x + 9)^{2}\).

Answer:

• \((2x - 3)(x + 5)=0\): \(x=\frac{3}{2},x=-5\)
• \((x + 3)(2x-1)=0\): \(x=-3,x=\frac{1}{2}\)
• \(2x^{2}+7 = 169\): \(x=\pm9\)
• \(5(x + 1)^{2}-2 = 43\): \(x = 2,x=-4\)
• \(x^{2}+x-30 = 0\): \(x=-6,x = 5\)
• \(x^{2}+5x-14 = 0\): \(x=-7,x = 2\)
• \((x - 7)^{2}=25\): \(x = 12,x = 2\)
• \(x^{2}+18x+\underline{}\): 81 (and \((x + 9)^{2}\))