McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
1. Solving Quadratic Equations by Factoring
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Exercise 44 Page 174

Using the given information, write the lengths of the sides and the hypotenuse in terms of each other. Then consider the Pythagorean Theorem to find the dimensions of the triangle.

7 cm, 24 cm, and 25 cm

Practice makes perfect

Let a and b be the lengths of the legs and c be the length of the hypotenuse.

Using the given information, let's show the relations between these lengths.

Verbal Expression Algebraic Expression
Hypotenuse is 1 cm longer than one side c= a+1
Hypotenuse is 4 cm longer than 3 times the other side c=3 b+4

To find the dimensions of the triangle, we can apply the Pythagorean Theorem. However, we have more than one unknowns in this situation. To reduce the unknowns, we should rewrite the lengths in terms of each other. c is already written in terms of b. Let's also write a in terms of b.

c=3b+4
a+1=3b+4
a=3b+3

Now we are ready to apply the theorem.

a^2+b^2=c^2
( 3b+3)^2+b^2=(3b+4)^2
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Simplify
9b^2+18b+9+b^2=9b^2+24b+16
18b+9+b^2=24b+16
-6b+9+b^2=16
-6b-7+b^2=0
b^2-6b-7=0
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Factorize
b^2+b-7b-7=0
b(b+1)-7b-7=0
b(b+1)-7(b+1)=0
(b+1)(b-7)=0
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Solve using the Zero Product Property
lb+1=0 b-7=0
lb=-1 b-7=0
lb=-1 b=7

Because the length cannot be negative, the length of one side is 7 cm. With this, we can find the length of the other side and the hypotenuse. a=3b+3 ⇒ a=3( 7)+3=24 c=3b+4 ⇒ c=3( 7)+4=25 Therefore, the dimensions of the triangle can be listed as 7 cm, 24 cm, and 25 cm.