All categories

3 years ago

Determine all values of k for which the roots of the equation x^2-6x+k=0 are complex

Mathematics

1 answer:

MaRussiya [10]3 years ago

6 0

Complex roots occur when b^2 - 4ac is negative.

So we have (-6)^2 - 4.1.k < 0
-4k < - 36
k > 9

You might be interested in

Simplify the expression show all your work 18 / 2 x 3 over 5 - 2

11Alexandr11 [23.1K]

Answer:

18÷2 × 3 / 5-2

9 ×3 / 3

27/3

tyvm this was my 100th answer :)

6 0

3 years ago

PLEASE HELP!!!! 20 POINTS!!!! What best describes the following equation?

TiliK225 [7]

It’s both a relation and a function

5 0

3 years ago

What do you know to be true about the values of p and q?

Harrizon [31]

The thing that's true about the values p and q is that p = q.

The total <em>sum of the angles</em> in a triangle is 180°.

From the first triangle, the value of p will be:

80° + 20° + p = 180°

100° + p = 180°

p = 180° - 100°

p = 80°

From the second triangle, the value of q will be:

55° + 45° + q = 180°

100° + q = 180°

q = 180° - 100°

q = 80°

Therefore, p = q.

Read related link on:

brainly.com/question/16020981

7 0

3 years ago

Solve the following equation by factoring:9x^2-3x-2=0

olya-2409 [2.1K]

Answer:

The two roots of the quadratic equation are

$x_1= - \frac{1}{3} \text{ and } x_2= \frac{2}{3}$

Step-by-step explanation:

Original quadratic equation is $9x^{2}-3x-2=0$

Divide both sides by 9:

$x^{2} - \frac{x}{3} - \frac{2}{9}=0$

Add $\frac{2}{9}$ to both sides to get rid of the constant on the LHS

$x^{2} - \frac{x}{3} - \frac{2}{9}+\frac{2}{9}=\frac{2}{9}$ ==> $x^{2} - \frac{x}{3}=\frac{2}{9}$

Add $\frac{1}{36}$ to both sides

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{2}{9} +\frac{1}{36}$

This simplifies to

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{1}{4}$

Noting that (a + b)² = a² + 2ab + b²

If we set a = x and b = $\frac{1}{6}\right)$ we can see that

$\left(x - \frac{1}{6}\right)^2$ = $x^2 - 2.x. (-\frac{1}{6}) + \frac{1}{36} = x^{2} - \frac{x}{3}+\frac{1}{36}$

$\left(x - \frac{1}{6}\right)^2=\frac{1}{4}$

Taking square roots on both sides

$\left(x - \frac{1}{6}\right)^2= \pm\frac{1}{4}$

So the two roots or solutions of the equation are

$x - \frac{1}{6}=-\sqrt{\frac{1}{4}}$ and $x - \frac{1}{6}=\sqrt{\frac{1}{4}}$

$\sqrt{\frac{1}{4}} = \frac{1}{2}$

So the two roots are

$x_1=\frac{1}{6} - \frac{1}{2} = -\frac{1}{3}$

and

$x_2=\frac{1}{6} + \frac{1}{2} = \frac{2}{3}$

7 0

2 years ago

BaLLatris [955]

Answer:

160 is the answer please tell me if im wrong

5 0

2 years ago