Question

For what value of k does 4x2-12x-k =0 have no real roots

Answer 1

$\huge\bf\mathscr\pink{Your\: Answer}$

K < -9

step-by-step explanation:

Given quadratic equation,

4x^2 - 12x - K = 0

According to the general form of a quadratic equation,

ax^2 + bx + c = 0

Here,

a = 4

b = -12

c = -K

Now,

we know that,

In a quadratic equation,

There will be no real roots only when the

Descriminant ( D) < 0

i.e.,

Descriminant is neagative.

A.T.Q

D < 0

=> b^2 - 4ac < 0

Putting the values of a, b and c

we get,


=> (-12)^2 - 4 × (4)×(-K) < 0

=> 144 + 16K < 0

=> 16K < -144

=> K < -144/16

=> K < -9

Hence, for no real roots,

the value of

K < -9

Answer 2

any quadratic equation, $ax^2+bx+c=0$ doesn't have real roots only when discriminant, $\bf{D=b^2-4ac<0}$

here quadratic equation is 4x² - 12x - k = 0 doesn't have real roots.

so, D = (-12)² - 4(4)(-k) < 0

or, 144 + 16k < 0

or, 16 × 9 + 16k < 0

or, 16(9 + k) < 0

or, 9 + k < 0

or, k < -9

hence, for $k\in(-\infty,-9)$ does 4x² - 12x - k have no real roots .