Chapter A.4, Problem 31E

### Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

Chapter
Section

### Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem

# In Exercises 25 to 32, solve each quadratic equation by factoring. 6 x 2 = 5 x - 1

To determine

To solve:

6x2=5x-1

Explanation

Solving an equation is to find the value of the unknown variables in the equation, such that 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,

6x2=5x-1

The above equation is not in the standard form of a quadratic equation.

Add, -5x+1 on the both side

6x2+-5x+1=5x-1+-5x+1

6x2-5x+1=5x-5x+-1+1

6x2-5x+1=0

Now, we shall proceed with the factorization of the quadratic expression in the left of the.

First look for the common factor GCF of the terms 6x2, -5x and 1.

But, here the GCF is 1.

Now, factor the expression 6x2-5x+1 by the reverse FOIL method.

First factorize the product of the coefficient of the first tern and the constant, such that the sum of the factors gives the coefficient of the middle term.

Thus, factorize 6 as -3Â·2 so that the sum o f the factors -3+(-2) gives -5.

Now, split the middle term -5x as (-3x)+(-2x).

Hence, the given equation becomes,

6x2+-3x+-2x+1=0

3xÂ·2x+3xÂ·-1+-1Â·2x+-1-1=0

Factor out the common factors form the first two terms and form the last two terms of the above expression

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