The browser that is used for submitting questions does not permit many mathematical symbols. It is therefore not possible to be sure what the question is. For a quadratic equation of the form y = ax^2 + bx + c, where a, b and c are real numbers and a is non-zero, the discriminant is b^2 – 4ac.
Discriminant = b2-4ac = (-2)2-(4*4*4) = -60
16x = 4 Divide both sides by '16' Hence 16x/16 = 4/16 x = 4 / 16 x = 1/4
It has 2 real solutions because the discriminant is greater than zero.
Divide by 16, which gives you x=4/16 The 4/16 cancels down to 1/4.
16x/6
If your function is 4x^2+16x+16=0 then x=-2, if it is 4x^2-16x+16=0, then x=2
It is; (2x+4)^2.
4(x + 2)(x + 2)
The discriminant is 0.
The discriminant is 33.
4x2(4x3 + 3)
True
4x2-11x-20=(x-4)(4x+5) 4x2-11x-20=4x2+ax+bx-20ab = 4*-20 = -80a+b = -11a = -16, b = 54x2-11x-20=4x2-16x+5x-204x2-16x+5x-20=4x(x-4)+5(x-4)4x(x-4)+5(x-4)=(x-4)(4x+5)Factorised, 4x2-11x-20=(x-4)(4x+5)
8-16x-32=0 8=16x-32 2x-4=0 2x=4 x=2
x3 + 4x2 + 16x + 64 =x2*(x + 4) + 16(x + 4) = (x + 4)*(x2 + 16) which has no further real factors.
4x2 - 16x + 12
zero