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Q. 70

Expert-verified
Found in: Page 210

Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

In Problems 63–72, find the real solutions of each equation.${x}^{4}-2{x}^{3}+10{x}^{2}-18x+9=0$

The real solution of the equation is $1.$

See the step by step solution

Step 1. Finding the possible number of zeros

The given equation is ${x}^{4}-2{x}^{3}+10{x}^{2}-18x+9=0$

The given function is of degree four, so it has at most four real zeros.

Step 2. Use the rational zero theorem

As all the coefficients are integers so we use the rational zeros theorem.

The factors of the constant term $9$ are

$p:±1,±3,±9$

The factors of the leading coefficient $1$ are

$q:±1$

So, the possible rational zeros are

$\frac{p}{q}:±1,±3,±9$

Step 3. Finding the zero

Let's test the potential zero $1$ by using substitution

Thus,

$f\left(1\right)={\left(1\right)}^{4}-2{\left(1\right)}^{3}+10{\left(1\right)}^{2}-18\left(1\right)+9\phantom{\rule{0ex}{0ex}}f\left(1\right)=0$

Since the remainder is zero, $\left(x-1\right)$ is a factor.

Step 4. Use synthetic division

Use synthetic division to factor equation,

Step 5. Use the zero product property

As the remaining zeros satisfy the depressed equation

So,

${x}^{3}-{x}^{2}+9x-9=0\phantom{\rule{0ex}{0ex}}{x}^{2}\left(x-1\right)+9\left(x-1\right)=0\phantom{\rule{0ex}{0ex}}\left({x}^{2}+9\right)\left(x-1\right)=0$

Use the property

${x}^{2}+9=0\phantom{\rule{0ex}{0ex}}{x}^{2}=-9$ and $x-1=0\phantom{\rule{0ex}{0ex}}x=1$

The square of a negative number is imaginary thus. it is not real.