Core Connections Geometry, 2013
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Core Connections Geometry, 2013 View details
1. Section 5.1
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Exercise 29 Page 290

Practice makes perfect
a To solve the equation, we must first eliminate the coefficient to x^2 by dividing both sides by 6. Then we can take the square root of both sides to isolate x.
6x^2=150
x^2=25
x=± 5
We have two solutions, x=- 5 and x=5. To check our solutions we substitute them into the original equation. If the left-hand side and right-hand side are equal, the solutions are correct. Let's test the first solution.
6x^2=150
6( - 5)^2 ? = 150
â–Ľ
Simplify
6(5)^2 ? = 150
6(25)? =150
150=150
Let's also test the second solution.
6x^2=150
6( 5)^2 ? = 150
â–Ľ
Simplify
6(25)? =150
150=150
Both solutions are correct.
b Before we can solve this equation we have to distribute the factor outside of the parentheses on the right-hand side. Then we can perform inverse operations until m is isolated.
4m+3-m=3(m+1)
3m+3=3(m+1)
3m+3=3m+3
3=3
The equation simplifies to an identity, which means the solution to this equation is all real numbers.
c Since the inverse operation to a square root is to square, we have to square both sides to eliminate the square root.
sqrt(5x-1)=3
(sqrt(5x-1))^2=3^2
5x-1=3^2
â–Ľ
Solve for x
5x-1=9
5x=10
x=2
Note that squaring an equation can possibly introduce false roots. Therefore, we must test our solution by substituting it into the original equation and then simplify.
sqrt(5x-1)=3
sqrt(5( 2)-1)? =3
â–Ľ
Simplify
sqrt(10-1)? =3
sqrt(9)? =3
3=3
The solution is correct.
d In this equation we have the square of an expression on the left-hand side and a negative number on the right-hand side. Since the square of something is always positive, the left-hand side can never equal - 3 no matter the value of k.

(k-4)^2 ≠ - 3 Therefore, there are no real solutions to this equation.