2.13 warm up determine whether the given value is a solution of the equation. 1.x² - 2x + 15 = 0; 3...
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2.13 Warm UpDetermine whether the given value is a
solution of the equation.
1. x² - 2x + 15 = 0; 3
2. x² + 3x – 4 = 0; 1
3. x² - 4x – 12 = 0; 2
2.13 Using Square Roots to Solve Quadratic Equations
• All positive real numbers have 2 square rootsPositive square root (principle square root)
Negative square root
• Radicand: number or expression inside the radical symbol
• Perfect square: squared integer
• If there is no b, you can isolate the x2 and then solve using square roots.
EXAMPLE 1 Solve quadratic equations
Solve the equation.
a. 2x2 = 8
SOLUTION
a. 2x2 = 8 Write original equation.
x2 = 4 Divide each side by 2.
x = ± 4 = ± 2 Take square roots of each side. Simplify.
ANSWER The solutions are –2 and 2.
Solve quadratic equationsEXAMPLE 1
b. m2 – 18 = – 18 Write original equation.
m2 = 0 Add 18 to each side.
The square root of 0 is 0.m = 0
ANSWER
The solution is 0.
Solve quadratic equationsEXAMPLE 1
c. b2 + 12 = 5 Write original equation.
b2 = – 7 Subtract 12 from each side.
ANSWER
Negative real numbers do not have real square roots. So, there is no real solution.
EXAMPLE 2 Take square roots of a fraction
Solve 4z2 = 9.
SOLUTION
4z2 = 9 Write original equation.
z2 = 94 Divide each side by 4.
Take square roots of each side.z = ± 94
z = ± 32
Simplify.
The solutions are – and 32
32
Approximate solutions of a quadratic equation
EXAMPLE 3
Solve 3x2 – 11 = 7. Round the solutions to the nearesthundredth.
SOLUTION
3x2 – 11 = 7 Write original equation.
3x2 = 18 Add 11 to each side.
x2 = 6 Divide each side by 3.
x = ± 6 Take square roots of each side.
The solutions are about – 2.45 and about 2.45.
EXAMPLE 1 Solve quadratic equations
Solve the equation.
1. c2 – 25 = 0
GUIDED PRACTICE for Examples 1,2 and 3
ANSWER –5, 5.
2. 5w2 + 12 = – 8 ANSWER no solution
3. 2x2 + 11 = 11 ANSWER 0
EXAMPLE 1 Solve quadratic equations
Solve the equation.
4. 25x2 = 16
GUIDED PRACTICE for Examples 1,2 and 3
ANSWER 4 5
4 5– ,
5. 9m2 = 100 ANSWER 103– ,
103
6. 49b2 + 64 = 0 ANSWER no solution
Solve the equation. Round the solutions to the nearest hundredth.
7. x2 + 4 = 14 ANSWER – 3.16, 3.16
8. 3k2 – 1 = 0 ANSWER – 0.58, 0.58
9. 2p2 – 7 = 2ANSWER – 2.12, 2.12
EXAMPLE 4 Solve a quadratic equation
Solve 6(x – 4)2 = 42. Round the solutions to the nearesthundredth.
6(x – 4)2 = 42 Write original equation.
(x – 4)2 = 7 Divide each side by 6.
x – 4 = ± 7 Take square roots of each side.
7 x = 4 ± Add 4 to each side.
ANSWER
The solutions are 4 + 6.65 and 4 – 1.35.7 7
EXAMPLE 1 Solve quadratic equations
Solve the equation. Round the solution to the nearest hundredth if necessary.
10. 2(x – 2)2 = 18
GUIDED PRACTICE for Examples 4 and 5
ANSWER –1, 5
11. 4(q – 3)2 = 28 ANSWER 0.35, 5.65
12. 3(t + 5)2 = 24 ANSWER –7.83, –2.17