Question

PLZ FAST Solve for x in the equation 2 x squared minus 5 x + 1 = 3. x = five-halves plus-or-minus StartFraction StartRoot 29 EndRoot Over 2 EndFraction x = five-halves plus-or-minus StartFraction StartRoot 41 EndRoot Over 4 EndFraction x = five-fourths plus-or-minus StartFraction StartRoot 29 EndRoot Over 2 EndFraction x = five-fourths plus-or-minus StartFraction StartRoot 41 EndRoot Over 4 EndFraction

Answer 1

Option D

Explanation:

Took the test, its correct

Answer 2

`(5)/(4) +/-(√(41) )/(4)`

Explanation:

We re-write the equation subtracting from both sides 3, so as to get a quadratic expression equal to zero and be able to use the quadratic formula to solve it:

`2x^2-5x+1=3\\2x^2-5x+1-3=0\\2x^2-5x-2=0`

Now we use the quadratic formula considering that
`a=2, b=-5,\,and\,\,c=-2`

`x=(-b+/-√(b^2-4\,a\,c) )/(2\,a) \\x=(-(-5)+/-√((-5)^2-4\,(2)\,(-2)) )/(2\,(2)) \\\\x=(5+/-√(25+16) )/(4) \\x=(5+/-√(41) )/(4) \\x=(5)/(4) +/-(√(41) )/(4)`