Use factoring to solve the equation. 15) 3x^2 – 5x – 18 = 0 A) -2, -3 B) 2, 3 C) 3, 2, 3 D) 0, 2, 3 Solve by the square-root property: 16) (x + 4)^2 = 20 A) 2 – √15, -5 B) 4 – 2√15, 4 + 2√15 C) 2 – √45, 2 + √45 D) 4 + 2√16, -4 – 2√16
The Correct Answer and Explanation is:
Let’s tackle both problems step by step using the appropriate algebraic methods.
Problem 15: Factor the quadratic equation Given: 3×2−5x−18=03x^2 – 5x – 18 = 0
We begin by looking for two numbers whose product is 3×−18=−543 \times -18 = -54 and whose sum is −5-5. These numbers are −9-9 and 66.
Rewrite the middle term: 3×2−9x+6x−18=03x^2 – 9x + 6x – 18 = 0
Group terms: (3×2−9x)+(6x−18)=0(3x^2 – 9x) + (6x – 18) = 0
Factor each group: 3x(x−3)+6(x−3)=03x(x – 3) + 6(x – 3) = 0
Now factor the common binomial: (3x+6)(x−3)=0(3x + 6)(x – 3) = 0
Divide out the constant: 3(x+2)(x−3)=03(x + 2)(x – 3) = 0
Set each factor equal to zero: x+2=0x + 2 = 0 gives x=−2x = -2 x−3=0x – 3 = 0 gives x=3x = 3
Answer: A) -2, -3
Problem 16: Solve using the square-root property Given: (x+4)2=20(x + 4)^2 = 20
Apply the square-root property: x+4=±20x + 4 = \pm \sqrt{20}
Simplify the radical: 20=4⋅5=25\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5} So: x+4=±25x + 4 = \pm 2\sqrt{5}
Now isolate x: x=−4±25x = -4 \pm 2\sqrt{5}
Thus, the two solutions are: x=−4+25x = -4 + 2\sqrt{5} and x=−4−25x = -4 – 2\sqrt{5}
Answer: B) -4 + 2√5, -4 – 2√5
Both problems demonstrate classic quadratic solution methods. Factoring rewrites the equation into a product of binomials to extract roots directly, while the square-root property isolates a squared term and solves through root manipulation. These skills are essential for mastering algebraic analysis.
