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McGraw Hill Integrated II, 2012

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McGraw Hill Integrated II, 2012 View details

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Exercise 10 Page 803

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

± 9

Practice makes perfect

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

a^2+40^2=41^2

Calculate power

a^2+1600=1681

LHS-1600=RHS-1600

a^2 = 81

sqrt(LHS)=sqrt(RHS)

sqrt(a^2)=sqrt(81)

sqrt(a^2)=± a

a = ±sqrt(81)

Calculate root

a=± 9

We found that a=± 9. Therefore, there are two solutions for the equation, which are a=9 and a=- 9.

Checking Our Answer

Checking our answer

We can check our answers by substituting them for a in the given equation. Let's start with a=- 9.

a^2+40^2=41^2

a= - 9

( - 9)^2+40^2? =41^2

▼

Simplify

NegBaseToPosPow

(- a)^2 = a^2

9^2+40^2? =41^2

Calculate power

81+1600? =1681

Add terms

1681=1681 ✓

Since we got a true statement, we know that a=- 9 is a solution of the equation. Let's check if a=9 is also a solution.

a^2+40^2=41^2

a= 9

9^2+41^2? =41^2

▼

Simplify

Calculate power

81+1600? =1681

Add terms

1681=1681 ✓

We obtained a true statement, so we know that a=9 is also a solution of the equation.

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