Chapter 7.7, Problem 38WE

a.

To determine

Toshow: that $\frac{5-i\sqrt{2}}{4}$ is a root of $16x^2-40x+27=0$ .

Explanation of Solution

Given information:

The quadratic equation is $16x^2-40x+27=0$ .

Proof: if $\frac{5-i\sqrt{2}}{4}$ is a root of $16x^2-40x+27=0$ then after substitute the root in the equation, it satisfy the equation,

$\begin{array}{l}16x^2-40x+27=0\\ 16{\left(\frac{5-i\sqrt{2}}{4}\right)}^2-40\left(\frac{5-i\sqrt{2}}{4}\right)+27=0\text{ [substitute]}\\ 16\left(\frac{25+2i^2-10i\sqrt{2}}{16}\right)-50+10\sqrt{2}i+27=0\text{ [expand square and multiply]}\\ 25-2-50+27=0\text{ [}{i}^2=-1\text{ and cancel out common]}\\ 0=0\text{ [simplify]}\end{array}$

b.

To determine

To find: the another root.

Answer to Problem 38WE

The another root is $\frac{5+i\sqrt{2}}{4}$ .

Explanation of Solution

Given information:

The quadratic equation is $16x^2-40x+27=0$ .

Calculation:it is known that if a quadratic equation has a imaginary root then the conjugate of imaginary root must be a root of that quadratic equation.

Thus, the another root of quadratic equation is $\frac{5+i\sqrt{2}}{4}$ .