how to find solutions?

Question

can someone explain to me how you would find the solutions for this equation?


x^2-8x+3=0

Answer 1

*Since it can't be factored, use the quadratic formula...
x = [-b +/- √b^2 - 4ac] / 2a

First: a = 1, b = -8 & c = 3 > substitute the numbers with the corresponding variables.

x = [-(-8) +/- √(-8)^2 - 4(1)(3)] / 2(1)
x = [8 +/- √(-8)(-8) - 4(1)(3)] / 2
x = [8 +/- √64 - 4(3)] / 2
x = [8 +/- √64 - 12] / 2
x = [8 +/- √52] / 2
x = [8 +/- √2*2*13] / 2

Sec: when a number is repeated twice (in a radical sign), place it (once) in front of its radical sign.

x = [8 +/- 2√13] / 2
x = 8/2 +/- (2/2)√13
x = 4 +/- √13

Answer 2

x^2 - 8x + 3 = 0
Subtract 3 from both sides :
x^2 - 8x = -3
Calculate half of 8 (= 4), then square it (= 16).
Now add this 16 to both sides :
x^2 - 8x + 16 = -3 + 16 = 13
Factorise the left-hand side, as it is now a square :
(x - 4)^2 = 13
Take the square root of both sides :
x - 4 = ± sqrt(13)
Add 4 to both sides :
x = 4 ± sqrt(13), which are the 2 solutions to the equation.

Or you could use the quadratic formula :
x = [-b ± sqrt(b^2 - 4ac)] / (2a), where a = 1, b = -8, c = 3.
I'll let you do the calculation.

Answer 3

[ 8 ± √(64 - 12) ] / 2
[ 8 ± √(52) ] / 2
[ 8 ± 2√(13)] / 2
4 ± √(13)

Answer 4

no rational roots. Use quadratic formula

x= [- b±√(b²-4ac)]/2a
x= 8±√(64-4(1*3) /2
x=4±√13