Solve 2x^2-8x-3=0 by completing the square?

Question

Show working out, but no need to make it too detailed.

Answer 1

2x^2 - 8x - 3 = 0

Divide out the 2

x^2 - 4x - 3/2 = 0

Move constant to the other side

x^2 - 4x = 3/2

Half and square -4 and add to both sides

x^2 - 4x + 4 = 3/2 + 4

Write left side as a square, simplify the right

(x - 2)^2 = 22/4

Take the square root

x - 2 = +/- SQRT(22) / 2

Add 2

x = 2 +/- SQRT(22) / 2

Answer 2

Given above equation,
That is:

2x^2-8x-3=0

Make the coefficient of x^2 unitary by dividing by 2

We have:
x^2-4x-3/2 = 0
x^2-4x= 3/2

Half coefficient of x square and add to both sides.

x^2 - 4x + (-4/2)^2 = 3/2 + (-4/2)^2
x^2 - 4x + (-2)^2 = 3/2 + (-2)^2
x^2 - 4x + 4 = 3/2 + 4
(x-2)^2 = 11/2
x = 2 + (11/2))^0.5 = 2 + 2.34520788 = 4.34520788

OR

x = 2 - (11/2))^0.5 = - 0.3452078799

Answer 3

2x^2 - 8x - 3 = 0

x^2 - 4x - 3/2 = 0

x^2- 4x + 4 = 3/2 + 4

(x-2)^2 = 11/2

(x-2) = sqrt (11/2)

x = 2 +/- sqrt (11/2)

or x = 2 +/- sqrt (22/4)

x = 2 +/- sqrt (22) /2

Answer 4

so this comes like this.

write this equation as this:

2x^2-8x+8-11 = 0

2(x^2-4x+4) = 11

(x-2)^2 = 11/2

x = (+)(-)sqrt (6.5) +2

Answer 5

2.(x² - 4x - 3/2) = 0
x² - 4x - 3/2 = 0
(x² - 4x + 4) - 4 - 3/2 = 0
(x - 2)² = 11/2
x - 2 = ± √(11/2)
x = 2 ± √(11/2)

Answer 6

u use the quadratic formula with 2 as a, -8 as b and -3 as c
then solve the answers u get are the zero's for the equation then just use that to create the equation

Answer 7

divide the equation by 2
x^2-4x-3/2=0
x^2-2*(2)(x)-3/2=0
x^2-2*(2)(x)+2^2 -4-3/2=0
x^2-2*(2)(x)+2^2-11/2=0
(x+2)^2-[sqrt(11/2)]^2=0
[(x+2+sqrt(11/2)][x+2-sqrt(11/2)]=0 ... a^2-b^2=(a+b)(a-b)
Either x+2+sqrt(11/2)=0--> x=-2-sqrt(11/2)
or x+2-sqrt(11/2)=0--> x=-2+sqrt(11/2)

Answer 8

the equation is of the form ax^2+bx+c=0

the formula to be used here is

-b (+ or -) [{sqrt(b^2-(4*a*c))}/2a]
we have a=2,b= -8, c= -3
= 8(+ or -) [{sqrt(64 + 24)}/4]
=8( + or -) [ 9.38/4]
= 8 (+ or - ) (2.345)
x = 10.345 or 5.655