Question

If the roots of x²+kx+k+3=0 are real and equal, what is the value of k?

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Answer 1

Answer:

Hey There!

According to the question,of x²+kx+k+3=0 are real and equal.

So here we have to find out the value of k..

We know that general form of quadratic equation is

${ax}^{2} + bc + c$

Now on comparing ax^2+bc+c with

${x}^{2} + kx + k = 0 \: we \: get \: \: \\ a = 1 \: b = k \: and \: c =k$

We know that the roots are equal here....

so let's find out what is the discrimant here...

${b}^{2} - 4ac = 0 \\ \to {k}^{2} - 4k = 0 \\ \to \: k(k - 4) = 0 \\ \to discrimant = 0$

So k=0,4

Hence value of k is 0,4

I hope it is helpful to you...

Cheers!___________

Answer 2

The values of k if the roots of x²+kx+k+3=0 are real and equal, are 6 and -2

The quadratic equation is given as:

$x^2 + kx + k + 3 = 0$

A quadratic equation is represented as:

$ax^2 + bx + c = 0$

By comparison, we have:

$a = 1$

$b = k$

$c = k + 3$

When the roots are real and equal, then:

$b^2 = 4ac$

So, we have:

$k^2 = 4 * 1 * ( k + 3)$

$k^2 = 4k + 12$

Rewrite as:

$k^2 - 4k - 12 = 0$

Expand

$k^2 +2k- 6k - 12 = 0$

Factorize

$k(k +2)- 6(k + 2) = 0$

Factor out k + 2

$(k - 6)(k + 2) = 0$

Solve for k

k = 6 or -2

Hence, the values of k are 6 and -2

Read more about quadratic functions at:

https://brainly.com/question/1770447