BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

Solutions

Chapter 1.5, Problem 89E
To determine

Find the real solutions of equation.

Expert Solution

Answer to Problem 89E

  x=2,x=4

Explanation of Solution

Given information:

  x+5x2=5x+2+28x24

Formula used: -

First, we will solve this equation by cross multiplication method and then we will find value of x.

  x=b±b24ac2a

  x+5x2=5x+2+28x24

Now converting (x24) to (x+2)(x2)

  x+5x2=5x+2+28(x+2)(x2)

Now multiplying (x+2)(x2) to each and every term of equation:

  (x+2)(x2)x+5x2=(x+2)(x2)5x+2+(x+2)(x2)28(x+2)(x2)=(x+2)(x+5)=5(x2)+28=x2+5x+2x+10=5x10+28x2+2x8=0

Now applying quadratic formula:

  x=b±b24ac2a In equation x2+2x8=0

Here a=1,b=2,c=8

  x=(2)±(2)24(1)(8)2(1)x=2±4+322x=2±362x=2±62

  x=2+62,x=262x=2,x=4

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