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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 + x 2 1 = 0

To determine

To calculate: The all real solutions in the polynomial equation x4+x21=0.

Explanation

Given Information:

The provided polynomial equation is x4+x21=0.

Formula used:

The value of x in the equation ax2+bx+c=0 can be given by the quadratic formula,

x=b±b24ac2a where a,b and c are real numbers with a0.

Calculation:

Consider the polynomial equation x4+x21=0.

The above polynomial equation can be written as (x2)2+x21=0

Let x2=t put in the polynomial equation (x2)2+x21=0

t2+t1=0

Compare the above equation with the equation ax2+bx+c=0, a=1, b=1 and c=1

Put the values of a,b,c in the quadratic formula x=b±b24ac2a.

t=1±(1)241(1)21=1±1+42=1±52

Further simplify,

Put 1±52 in x2=t, x2=t=1±52

Take square root from both sides,

x=±1±52

Check:

For x=1±52

Substitute x=1±52 in the equation x4+x21=0 then simplify,

(1±52)4+(1±52)21=?0(1±52)2+(1±52)1=?0

Taking positive (+) sign from the above equation then simplify,

(1+52)2+(1+52)1=?01+5254+1+521=?06252+2544=?00=0

Which is true

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