Question

3. Determine k so that the given equation will have the stated property, and write the resulting equation:
(a) x
2 + 4kx + k + 2 = 0 has one root

Answer 1

Answer:

The values of k are

$k=\frac{1+\sqrt{33}}{8}$

$k=\frac{1-\sqrt{33}}{8}$

Step-by-step explanation:

we know that

In a quadratic equation of the form

$ax^{2} +bx+c=0$

The discriminant D is equal to

$D=b^{2}-4(a)(c)$

If D=0 ----> The quadratic equation has only one real solution

If D>0 ----> The quadratic equation has two real solutions

If D<0 ----> The quadratic equation has no solutions (complex solutions)

in this problem we have

$x^{2} +4kx+k+2=0$  

so

$a=1\\b=4k\\c=(k+2)$

Find out the discriminant D

substitute the values

$D=(4k)^{2}-4(1)(k+2)$

$D=16k^{2}-4k-8$

For D=0

$16k^{2}-4k-8=0$

Solve for k

$16k^{2}-4k=8$

Factor 16 left side

$16(k^{2}-\frac{1}{4}k)=8$

Complete the square

$16(k^{2}-\frac{1}{4}k+\frac{1}{64})=8+\frac{1}{4}$

$16(k^{2}-\frac{1}{4}k+\frac{1}{64})=\frac{33}{4}$

$(k^{2}-\frac{1}{4}k+\frac{1}{64})=\frac{33}{64}$

Rewrite as perfect squares

$(k-\frac{1}{8})^{2}=\frac{33}{64}$

square root both sides

$k-\frac{1}{8}=(+/-)\frac{\sqrt{33}}{8}$

$k=\frac{1}{8}(+/-)\frac{\sqrt{33}}{8}$

$k=\frac{1}{8}(+)\frac{\sqrt{33}}{8}=\frac{1+\sqrt{33}}{8}$

$k=\frac{1}{8}(-)\frac{\sqrt{33}}{8}=\frac{1-\sqrt{33}}{8}$

therefore

The values of k are

$k=\frac{1+\sqrt{33}}{8}$

$k=\frac{1-\sqrt{33}}{8}$