Which equation have no real solution?
a. x^2+4x+16=0
b.4x^2+4x-24=0
c.5x^2+3x-1=0
d.2x^2-4x+4=0

To find out which equation has no real solutions, we need to calculate the discriminant for each of these given equations.

For calculating the discriminant, we need to first compare these equations with the general formula which is ax²+bx+c.

So, let's get started.

1) x² + 4x + 16 = 0

a=1, b=4, c=16

D = b²-4ac

= (4)² - 4(1)(16)

= 16-64

= -48

√D = √-48

2) 4x² + 4x - 24 = 0

a=4, b=4, c=-24

D = b²-4ac

= (4)² - 4(4)(-24)

= 16 - 16(-24)

= 16 + 384

= 400

√D = √400 = +20 or -20

3) 5x² + 3x - 1 = 0

a=5, b=3, c=-1

D = b²-4ac

= (3)² - 4(5)(-1)

= 9 + 20

= 29

√D = √29

4) 2x² - 4x + 4 = 0

a=2, b=-4, c=4

D = b²-4ac

= (-4)² - 4(2)(4)

= 16 - 32

= -16

√D = √-16

Now from all these above calculations, we can see that discriminant was negative in first equation and in last equation.

If D<0 then roots does not exist, as the square root can not contain a negative value or the equation does not have any real solutions.

Roots in such case can be calculated but those roots are known as imaginary roots, which is a higher concept.

So Final answer is,

Equation 1 => x² + 4x + 16 = 0

and

Equation 4 => 2x² - 4x + 4 = 0

has no real solutions.

ok

step-by-step explanation:

51-50

step-by-step explanation: