Suggested languages for you:

Americas

Europe

Q. 70

Expert-verifiedFound in: Page 210

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 $1.$

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.

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

The factors of the constant term $9$ are

$p:\pm 1,\pm 3,\pm 9$

The factors of the leading coefficient $1$ are

$q:\pm 1$

So, the possible rational zeros are

$\frac{p}{q}:\pm 1,\pm 3,\pm 9$

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, $(x-1)$ is a factor.

Use synthetic division to factor equation,

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.

94% of StudySmarter users get better grades.

Sign up for free