Question

Determine the type and number of solutions of 4x2 − 3x + 1 = 0. A two imaginary solutions B two real solutions C one real solution

Answer 1

two imaginary solutions

Explanation:

Answer 2

Option A

The equation
`4x^2 - 3x + 1 = 0` has two imaginary solutions

Solution:

Given that we have to determine the type and number of solutions of given quadratic equation

`4x^2 - 3x + 1 = 0`

`\text {For a quadratic equation } a x^(2)+b x+c=0, \text { where } a \\eq 0\\\\x=\frac{-b \pm \sqrt{b^(2)-4 a c}}{2 a}`

In a quadratic equation, the discriminant helps tell you the number of real solutions to a quadratic equation

In the case of a quadratic equation
`ax^2 + bx + c = 0`, the discriminant is
`b^2 -4ac`

`\text{If } b^(2)-4 a c=0 ;$ then the given quadratic equation has one real root\\\\If b^(2)-4 a c>0,$ then the given quadratic equation has two real roots`

If
`b^2-4ac<0`, then the given quadratic equation has two imaginal roots which are complex conjugates

Given quadratic equation is:

`4x^2 - 3x + 1 = 0`

Here a = 4 and b = -3 and c = 1

The discriminant is given as:

`b^2 - 4ac = (-3)^2 - 4(4)(1) = 9 - 16 = -7\\\\b^2 - 4ac = -7\\\\b^2-4ac < 0`

Therefore the given equation has two imaginary solutions