Question

Find the value of k for which equation 4x^2 + 8x + k = 0 has real and
equal roots. Also find the solution of the given equation.

Pls Solve this And ill mark you brainliest Pls

Answer 1

Answer:

a=4. b=8. c=k

b^2 - 4ac

8^2 - 4*4*k

64 - 16k =0

16k = -64

k = -64/16

k = -4

real root doesnot exist

Answer 2

Quadratic equation

A quadratic equation in a variable $x$ is an equation which is of the form $ax^2 + bx + c = 0$ where constants a, b and c are all real numbers and $a \neq 0.$

Here, to find the value of k for the given quadratic equation let's compare the given equation with $ax^2 + bx + c = 0.$

After that it is given that the equation has real and equal roots, it means $b^2 - 4ac \leq 0$ and it has real and equal roots.

Now, find the coefficient and constant of the given equation.

Comparing the given equation with the standard form of quadratic equation, we get:

$\qquad \: a = 4, \; b = 8, \: c = k$

Find the discriminate. Now it is given that the equation has real and equal roots. So,

$\implies b^2 - 4ac \leq 0$

$\implies (8)^2 - 4 \times 4 \times k \leq 0$

$\implies 64 - 4 \times 4 \times k \leq 0$

$\implies 64 - 16 \times k \leq 0$

$\implies 64 - 16k \leq 0$

$\implies k = \dfrac{-64}{16}$

$\implies \boxed{k = -4}$

Hence, the value of k for the given quadratic equation is -4.