Question

Which two values of x are roots of 4x^2-6x+1

Answer 1

The roots of quadratic equation are$\frac{-6+ \sqrt{20}}{8} and \frac{-6- \sqrt{20}}{8}$

The roots of quadratic equation $ax² + bx+c$is given as,

$x = \frac{ - b± \sqrt{ {b}^{2} - 4ac } }{2a}$

Given equation is,

$4x²-6x + 1$

Substituting all values after comparison in above formulas,

$x = \frac{ - 6± \sqrt{36 - 16} }{8}$

$x = - 6± \sqrt{20}$

Hence, the roots of quadratic equation are $\frac{-6+ \sqrt{20}}{8} and \frac{-6- \sqrt{20}}{8}$

Answer 2

x= 3+√5/4,3-√5/4

Step-by-step explanation:

1.In general, given ax^2+bx+c=0, there exists two solutions where:

x= -b+√b^2 - 4ac/2a, -b-√b^2-4ac/2a

2.In this case, a=4, b=-6 and c=1.

x=6+√(-6)^2-4×4/2×4, 6-√(-6)^2-4×4/2×4

3.Simplify.

x=6+2√5/8, 6-√5/8

4.Simplify solutions.

x=3+√5/4, 3-√5/4