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Pearson Algebra 1 Common Core, 2015

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Pearson Algebra 1 Common Core, 2015 View details

3. Solving Quadratic Equations

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Exercise 1 Page 563

Consider the positive and negative solutions when isolating a variable raised to the power of two.

± 5

Practice makes perfect

To solve the given equation by finding the square roots, we need to consider the positive and negative solutions.

x^2 - 25 = 0

LHS+25=RHS+25

x^2 = 25

sqrt(LHS)=sqrt(RHS)

sqrt(x^2) = sqrt(25)

sqrt(a^2)=± a

x=± sqrt(25)

Calculate root

x = ± 5

We found that x=± 5. Thus, there are two solutions for the equation, which are x=5 and x=- 5.

Checking Our Answer

Checking our answer

We can check our answers by substituting them for x in the given equation. Let's start with x=- 5.

x^2-25=0

x= - 5

( - 5)^2-25? =0

▼

Simplify

NegBaseToPosPow

(- a)^2 = a^2

5^2-25? =0

Calculate power

25-25? =0

Subtract term

0=0 ✓

Since 0=0, we know that x=- 5 is a solution of the equation. Let's check if x=5 is also a solution.

x^2-25=0

x= 5

5^2-25? =0

Calculate power

25-25? =0

Subtract term

0=0 ✓

Again, since 0=0, we know that x=5 is a solution of the equation.

Subchapter links

Lesson Check p.563

Practice and Problem-Solving Exercises p.564-566