Question

Use the quadratic formula to find both solutions to the quadratic equation given below 2x^2-3x+1=0

Answer 1

Answer:

$x_1=1\\x_2=\frac{1}{2} =0.5$

Step-by-step explanation:

Given a equation of the form:

$ax^2+bx+c=0$

The roots of this equation can be found using the quadratic formula which is given by:

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

In this case we have this equation:

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

So:

$a=2\\b=-3\\c=1$

Using the the quadratic equation :

$x= \frac{-(-3)\pm\sqrt{(-3)^{2}-4(2)(1) } }{2(2)} = \frac{3\pm\sqrt{9-8 } }{4}=\frac{3\pm 1}{4}$

Therefore the two roots would be:

$x_1=\frac{3+ 1}{4}=\frac{4}{4}= 1\\x_2=\frac{3- 1}{4}=\frac{2}{4}=\frac{1}{2}=0.5$

Answer 2

-2   +   -1   =   -3
(2x2 - 3x) + 1 = 0
(x - 1) • (2x - 1) = 0
x-1=0
x = 1