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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 11 Page 803

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

± 15

Practice makes perfect

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

8^2+b^2=17^2

Calculate power

64+b^2=289

LHS-64=RHS-64

b^2=225

sqrt(LHS)=sqrt(RHS)

sqrt(b^2)=sqrt(225)

sqrt(a^2)=± a

b=±sqrt(225)

Calculate root

b=± 15

We found that b=± 15. Therefore, there are two solutions for the equation, which are b=15 and b=- 15.

Checking Our Answer

Checking our answer

We can check our answers by substituting them for b in the given equation. Let's start with b=- 15.

8^2+b^2=17^2

b= - 15

8^2+( - 15)^2? =17^2

▼

Simplify

NegBaseToPosPow

(- a)^2 = a^2

8^2+15^2? =17^2

Calculate power

64+225? =289

Add terms

289=289 ✓

We obtained a true statement, so b=- 15 is a solution of the equation. Let's check if b=15 is also a solution.

8^2+b^2=17^2

b= 15

8^2+ 15^2? =17^2

▼

Simplify

Calculate power

64+225? =289

Add terms

289=289 ✓

Again we got a true statement, so we know that b=15 is a solution of the equation.

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