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question 1 how many real solutions does $4x^2 - 3x + 1 = 0$ have? ○ 0 r…

Question

question 1
how many real solutions does $4x^2 - 3x + 1 = 0$ have?
○ 0 real solutions but 2 complex solutions
○ 2 real solutions
○ 1 real solution

question 2
1 pts
if the discriminant is greater than 0, the quadratic equation will have
○ 2 real solutions
○ 2 complex solutions
○ 1 real solution

Explanation:

Question 1

Step1: Recall discriminant formula

For quadratic equation \(ax^2 + bx + c = 0\), discriminant \(D = b^2 - 4ac\). Here, \(a = 4\), \(b = -3\), \(c = 1\).

Step2: Calculate discriminant

\(D = (-3)^2 - 4\times4\times1 = 9 - 16 = -7\).

Step3: Analyze discriminant

Since \(D < 0\), quadratic has 0 real solutions and 2 complex solutions (by quadratic formula properties).

Brief Explanations

For a quadratic equation \(ax^2 + bx + c = 0\), discriminant \(D = b^2 - 4ac\). If \(D > 0\), the square root of \(D\) is real and non - zero, so the quadratic formula \(x=\frac{-b\pm\sqrt{D}}{2a}\) gives two distinct real solutions.

Answer:

0 real solutions but 2 complex solutions

Question 2