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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

2. Solving Quadratic Equations by Graphing

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Exercise 61 Page 110

Take square roots on both sides of the equation.

- 3 ± sqrt(5)

Practice makes perfect

We will solve the equation and then check the solutions.

Solving the Equation

Let's solve the equation by taking square roots on both sides and then simplifying.

(x+3)^2=5

sqrt(LHS)=sqrt(RHS)

sqrt((x+3)^2)=sqrt(5)

sqrt(a^2)=± a

x+3=± sqrt(5)

LHS-3=RHS-3

x=- 3 ± sqrt(5)

Therefore, the solutions are - 3 + sqrt(5) and - 3 - sqrt(5).

Checking the Solutions

We can check the solutions by substituting them for x in the given equation. Let's start with - 3 + sqrt(5).

(x+3)^2=5

x= - 3 + sqrt(5)

( - 3 + sqrt(5)+3)^2 ? =5

Add terms

(sqrt(5))^2 ? =5

( sqrt(a) )^2 = a

5=5 ✓

Since substituting and simplifying created a true statement, we know that - 3 + sqrt(5) is a solution to the equation. Let's check if - 3 - sqrt(5) is also a solution.

(x+3)^2=5

x= - 3 - sqrt(5)

( - 3 - sqrt(5)+3)^2 ? =5

Add terms

(- sqrt(5) )^2 ? =5

NegBaseToPosPow

(- a)^2 = a^2

(sqrt(5))^2 ? =5

( sqrt(a) )^2 = a

5=5 ✓

We created another true statement, so we know that - 3 - sqrt(5) is also a solution to the equation.

Subchapter links

Exercises p.108-110