Find all values for K such that the equation has exactly one root HELP?

Question

How the hell do I solve this lol

Find all values of K such that the equation x^2+kx+79=0

has exactly one root!!

Please explain how you got your answer and i'll give you the best answer

Answer 1

This is a quadratic equation. It will have either two solutions or none. But it can have only one root if it is a double root. For example:

(x - a)² = 0 has a double root a. So you need to choose k so that the equation is a perfect square.

x² + kx + 79 = 0
(x - √79)² = 0 has a double root of +√79.

Expand it to find k.

(x - √79)² = x² - 2(√79)x + 79 = 0

So k = -2√79.

Answer 2

Use the discriminant: b^2 - 4ac.

If you have an equation in the form of ax^2 + bx + c = 0, it will have two unique solutions if the discriminant is greater than zero, one solution if the discriminant is equal to zero, and no solutions if it's less than zero.

So in the case of x^2 + kx + 79=0, we have a=1, b=k, and c=79. Plug this into the discriminant and you get k^2 - 4(1)(79), or k^2 - 316. So for the original equation to only have one root, this must equal zero. This is true when k = sqrt(316) or -sqrt(316).

The reason the discriminant works becomes clear if you look at the quadratic formula. The discriminant is the part found underneath the radical. If you can take the square root, then the "plus or minus" in the quadratic formula gives you two answers. If the discriminant is zero then the square root of it is zero, and the "plus or minus" zero in the quadratic formula only gives you the same answer twice. If the discriminant is less than zero, then you have no solution, because you can't take the square root of a negative number.

Answer 3

Find all values of K such that the equation x^2+kx+79=0.

(x + sqrt(79))^2 has [-sqrt(79)] as a root, of multiplicity 2.

Using FOIL: (x + sqrt(79))(x + sqrt (79)) = x^2 + 2sqrt(79)x + 79.
k = 2sqrt(79)

Answer 4

Dude, you should do your own homework but here's a pointer.

Check out: http://en.wikipedia.org/wiki/Quadratic_equation

See the formula for solving the quadratic equation. When the bit in the sqrt is zero then there will be only one answer.

So put b^2 - 4ac=0

So: k^2 -4*79 =0

k = sqrt(316)

Answer 5

well you need to do the same to both sides so...
x^2+kx+79=0
(then take x^2)
kx+79=-x^2
(then divide by x)
k+79=-x
(then take 79)
k=-79-x
and because there is 2 variables you cannot go ny thurther than this

by matt age 13

Answer 6

When the determinant b^2-4ac = 0, the equation has exactly one root.

k^2-4(79) = 0
k = ±2√79

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k can be positve or negative.

Answer 7

use quadratic formula