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Applied Calculus

7th Edition
Waner + 1 other
ISBN: 9781337291248

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Chapter
Section
BuyFindarrow_forward

Applied Calculus

7th Edition
Waner + 1 other
ISBN: 9781337291248
Textbook Problem

Solve the equations in Exercises 1–26.

2 ( x 2 1 ) x 2 + 1 x 4 x 2 + 1 x 2 + 1 = 0

To determine

To calculate: The solution of the equation 2(x21)x2+1x4x2+1x2+1=0.

Explanation

Given information:

The equation, 2(x21)x2+1x4x2+1x2+1=0.

Formula used:

Condition to solve an equation of the form PQ=0,

If PQ=0, then P=0.

Formula for difference of two squares,

(a+b)(ab)=a2b2

Where, a and b are real numbers.

Calculation:

Consider the expression,

2(x21)x2+1x4x2+1x2+1=0

The left side expression is 2(x21)x2+1x4x2+1x2+1.

Multiply both the numerator and the denominator by x2+1,

(2(x21)x2+1x4x2+1)x2+1(x2+1)x2+1=(2(x21)x2+1)x2+1(x4x2+1)x2+1(x2+1)x2+1=2(x21)(x2+1

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