   Chapter A.4, Problem 35E ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698

#### Solutions

Chapter
Section ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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# In Exercises 33 to 36, solve each incomplete quadratic equation. 4 y 2 = 25

To determine

To solve:

4y2=25.

Explanation

the obtained value or values of the unknown should satisfy the equation from which it was derived. Such a value is said to be the solution for the equation. In general a quadratic equation has two solutions for the variable in the equation as the degree of the equation is two.

Calculation:

Given,

4y2=25

First look for a common factor (GCF) of the terms 4y2, and 25.

But, here the GCF is 1.

Thus,

4y2=25

Add; -25 on both sides to make the equation to be in the standard form of a quadratic equation.

4y2+-25=25+-25

4y2-25=0

2·2y2-5·5=0

22y2-52=0

2y2-52=0

Now, factor the above equation by using the property, the difference of squares of a binomial a2-b2=a-ba+b

Here,

a=2y, and

b=5

Thus,

2y-52

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