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

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

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

$\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

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

Answer 2

Answer:

two imaginary solutions

Step-by-step explanation: