Question

the equation X square + 4 x + K is equals to zero has real roots find k​

Answer 1

Step-by-step explanation:

On comparing x^2+4x+k=0 with ax^2+bx+c=0 we get,. a=1,b=4,c=k

For real roots,

D> or =0

b^2-4ac> or =0

(4)^2-4(1)(k) > or = 0

16-4k > or = 0

4k> or = 16

k > or = 4.

Therefore, the value of k for which the equation has real roots is 4

Answer 2

Answer: $k\leq 4$

If the equation $x^{2} +4x+k = 0$ has real root then the value of $k$ must be less than or equal to 4.

Step-by-step explanation:

Given Quadratic equation is

$x^{2} +4x+k = 0$

If a quadratic equation $x^{2} +4x+k = 0$ has real roots. Therefore

$b^{2} -4ac\geq 0$ .........$(1)$

But We have $a$,$b$ and $c$ in given equation $x^{2} +4x+k = 0$ is

$a=1$ , $b = 4$ , $c = k$

Now, put these value in equation (1), We get,

$(4^{2})-4*1*k =\geq 0\\16-4k\geq 0\\16\geq 4k$

Dividing both sides by 4, We get

$4\geq k$

Therefore the value of $k$ must be less than or equal to 4 If equation $x^{2} +4x+k = 0$ has real root.