Question

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

Answer 1

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

Answer 2

`x_1=1\\x_2=(1)/(2) =0.5`

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)} = (3\pm√(9-8 ) )/(4)=(3\pm 1)/(4)`

Therefore the two roots would be:

`x_1=(3+ 1)/(4)=(4)/(4)= 1\\x_2=(3- 1)/(4)=(2)/(4)=(1)/(2)=0.5`