Question

How many real solutions does the equation 8x^2 − 10x + 15 = 0 have?

Answer 1

Answer:

The correct option is (A) No real solution.

Step-by-step explanation:

The expression provided is a quadratic equation.

$8x^{2}-10x+15=0$

The roots of a quadratic equation are:

$x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a }$

Here,

a = 8

b = -10

c = 15

The conditions to determine real and complex roots are:

• If $b^{2}-4ac$ then the quadratic equation has two complex roots.
• If $b^{2}-4ac>0$ then the quadratic equation has two real roots.

Compute the value of $b^{2}-4ac$ as follows:

$b^{2}-4ac=(-10)^{2}-(4\times 8\times15)$

             $=100-480\\=-380$

The equation has two complex roots.

Thus, the correct option is (A).