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antifrog16 · Oct 13, 2009, 01:31 PM

Hey could I have some help with these:

(b) 6 3 0 2
x − x + = by completing the square (giving an exact answer simplified if
possible). (3 marks)

(c) 2 7 3 2
x − x = by the formula (giving exact answers).

antifrog16 · Oct 13, 2009, 01:35 PM

Originally Posted by antifrog16

Hey could i have some help with these:

(b) 6 3 0 2
x − x + = by completing the square (giving an exact answer simplified if
possible). (3 marks)

(c) 2 7 3 2
x − x = by the formula (giving exact answers).

(b) x^2 - 6x + 3 = 0 by completing the square (giving an exact answer simplified if
possible). (3 marks)
(c) 2x^2 - 7x = 3 by the formula (giving exact answers). (2 marks)

sorry there are the questions

Unknown008 · Oct 14, 2009, 10:50 AM

For an equation $ax^2 + bx + c = 0$ , the completed square form is:

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

To do that, first of all remove (by factorisation) the coefficient of x^2:

$ax^2 + bx + c = 0$

$a(x^2 + \frac{bx}{a}) + c = 0$

Then, divide the coefficient of x by 2, replace x^2 by x and x by 1. Put all into brackets and square them. Also, subtract the square of the coefficient of x you just divided by 2:

$a[(x + \frac{b}{2a})^2 - (\frac{b}{2a})^2] + c = 0$

Then, re expand your values:

$a(x + \frac{b}{2a})^2 - (a)(\frac{b^2}{4a^2}) + c = 0$

$a(x + \frac{b}{2a})^2 - \frac{b^2}{4a} + c = 0$

Care to try the part (b) that you asked?

I (or another member) will tell you whether this is good.