Page 701 - Algebra 1

If no rounding instructions are given, round the approximation to the thousandths place.
Approximating Solutions
Example
3
Solve each equation.
a. x2=40 SOLUTION
2
√x = ± √4 · 1 0
2
√x = ± √4 · √1 0
x2 = 40 2
√ x = ± √ 4 0
Simplify the square root.
T a k e t h e s q u a r e r o o t o f b o t h s i d e s .
Find a factor that is a perfect square. Product Property of Radicals
S i m p l i f y .
x = ± 2 √ 1 0
Use a calculator to find the approximate value of √1 0.
Caution
Round after all computations have been made.
x ≈ 2 · (3.16227766) x ≈ ±6.32455532
x ≈ ±6.325
Write the approximate value. Multiply.
Round to the nearest thousandth.
x2 = 40 (-6.325)2 ≈Q 40
40.006 ≈ 40 ✓
Addition Property of Equality Combine like terms.
Division Property of Equality
Simplify.
Take the square root of both sides. Find the approximate square root. Round to the nearest thousandth.
8x2 -24=100 8(-3.937)2 - 24 ≈Q 100 8(15.499969) - 24 ≈Q 100 123.999752 - 24 ≈Q 100
✓ 99.999752 ≈ 100 ✓
Check x2 = 40 (6.325)2 ≈Q 40 40.006 ≈ 40
✓
b. 8x2 -24=100 SOLUTION
Begin by isolating x2. 8x2 -24=100
+_ 2 _ 4 +_ 2 _ 4
8x2 = 124
8x = 124 _2 _
88
Check 8x2 - 24 = 100 8(3.937)2 - 24 ≈Q 100 8(15.499969) - 24 ≈Q 100 123.999752 - 24 ≈Q 100 99.999752 ≈ 100
x2 = 15.5 2
√ x = ± √ 1 5 . 5
x ≈ ±3.937003937 x ≈ ±3.937
Caution
Remember to check both solutions.
686 Saxon Algebra 1