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Applied Calculus

7th Edition
Waner + 1 other
ISBN: 9781337291248

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Section
BuyFindarrow_forward

Applied Calculus

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

By any method, determine all possible real solutions of each equation in Exercises 13–30. Check your answers by substitution.

x 4 10 x 2 + 9 = 0

To determine

To calculate: The all real solutions in the polynomial equation x410x2+9=0.

Explanation

Given Information:

The provided polynomial equation is x410x2+9=0.

Formula used:

Principle of zero product rule:

An equation AB=0 is true if and only if A=0 or B=0, both.

A product is zero if and only if at least one factor is zero.

Calculation:

Consider the polynomial equation x410x2+9=0.

Factor the above equation,

x410x2+9=0x4(9+1)x2+9=0x49x2x2+9=0x2(x29)1(x29)=0(x29)(x21)=0

If (x29)=0 then,

Add 9 on both sides in the above equation,

x29+9=0+9x2+0=9x2=9

Take square root from both sides,

x2=9x=±3

If (x21)=0 then,

Add 1 on both sides in the above equation,

x21+1=0+1x2+0=1x2=1

Take square root from both sides,

x2=1x=±1

Check:

For x=1

Substitute the value of x=1 in x410x2+9=0.

x410x2+9=0(1)410(1)2+9=?0110+9=?00=0

Which is true

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