1. Completing the square

Hello,

I'm having problems with completing the square can anyone comment on my working?

2x^2 - 3x -1 = 0

2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0

2(x - 3/4)^2 -17/16 =0

2(x - 3/4)^2 = 17/16

(x - 3/4)^2 = 17/32

x - 3/4 = +-√17/32

x = 3/4 +-√17/32

2. Originally Posted by Mouseman
Hello,

I'm having problems with completing the square can anyone comment on my working?

2x^2 - 3x -1 = 0

2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0

2(x - 3/4)^2 -17/16 =0

2(x - 3/4)^2 = 17/16

(x - 3/4)^2 = 17/32

x - 3/4 = +-√17/32

x = 3/4 +-√17/32

2x^2 - 3x = 1

x^2 - (3/2)x = 1/2

(x - 3/4)^2 - 9/16 = 1/2

x - 3/4 = +-sqrt(17/16)

x = 3/4 +-sqrt(17/16)

3. Thank you very much!

4. Originally Posted by sean.1986
2x^2 - 3x = 1

x^2 - (3/2)x = 1/2

(x - 3/4)^2 - 9/16 = 1/2

x - 3/4 = +-sqrt(17/16)

x = 3/4 +-sqrt(17/16)
The answer is to be x = 3/4 +- √17/4

I can't see where I am going wrong.

5. When completing the square, just use the general formula.

$a(x+\frac{b}{2a})^{2}+c-\frac{b^{2}}{4a}$

Plug in a,b,c and you're set.

6. Second attempt

2x^2 - 3x -1 = 0

2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0

2((x - 3/4)^2) -17/8 =0

2(x - 3/4)^2 = 17/8

(x - 3/4)^2 = 17/16

x - 3/4 = +-√17/16

x = 3/4 +-√17/16

Is my teacher wrong in stating that it is x = 3/4 +-√17/4?

7. Just to add my 2 cents......

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

1. First transpose the -1 to the right side of the equation.

$2x^2-3x=1$

2. Divide each term by 2

$x^2-\frac{3}{2}x=\frac{1}{2}$

3. Take half of the coefficient of x, square it and add it to both sides.

$x^2-\frac{3}{2}x+(\frac{3}{4})^2=\frac{1}{2}+\frac{9}{ 16}$

4. Noting the perfect square trinomial on the left:

$(x-\frac{3}{4})^2=\frac{17}{16}$

5. Take the square root of both sides:

$x-\frac{3}{4}=\pm\sqrt\frac{17}{16}$

6. Finally,

$x=\frac{3}{4}\pm\frac{\sqrt17}{4}$

$x=\frac{3\pm\sqrt17}{4}$

8. Hello,

Originally Posted by Mouseman
Second attempt

2x^2 - 3x -1 = 0

2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0

2((x - 3/4)^2) -17/8 =0

2(x - 3/4)^2 = 17/8

(x - 3/4)^2 = 17/16

x - 3/4 = +-√17/16

x = 3/4 +-√17/16

Is my teacher wrong in stating that it is x = 3/4 +-√17/4?
Note that $16=4^2$

Therefore, $\sqrt{16}=4$, and this yields the result your teacher gave you

9. Yeah it's just a question of bracketing or simplifying.

sqrt (a/b) = sqrt(a) / sqrt(b)

so sqrt(17/16) = sqrt(17) / sqrt(16) = sqrt(17) / 4

Both answers are correct but I guess I should've simplified. Sorry mate!