# Rationalize denominator & factorising quadratic equations.

Tags:
1. Nov 1, 2015

### Meezus

1. The problem statement, all variables and given/known data

2. Relevant equations
Not Sure.

3. The attempt at a solution

For the first question I know you have to multiply the conjugate of the denominator so it would be (2 - √5)/(1−2√5) x (1−2√5)/(1−2√5) but I'm not sure how to actually do that.

For the second question. I have that -3 x 2 = -6 and -3 + 2 = -1 which fits the sum. I think this is correct.

for this second part of the second question, I'm not sure how to solve this as I don't see any factors of -6 which when added together = 11.

2. Nov 1, 2015

### andrewkirk

The first thing you need to do is to be very careful to get your signs right, which is not happening in the bit I quoted. Where has the 1+2√5 disappeared to?
Next, when multiplying the first fraction by $\frac{1-2\sqrt{5}}{1-2\sqrt{5}}$, just make the result a new fraction in which the numerator is the product of the numerators and the denominator is the product of the denominators. You know: $$\frac{a}{b}\times\frac{c}{d}=\frac{a\times c}{b\times d}$$.

3. Nov 1, 2015

### andrewkirk

You won't be able to use that simple rule, because the $x^2$ has a coefficient other than 1.

Instead set $(2x+a)(x+b)$ equal to the quadratic. Expand the brackets and then work out what $a$ and $b$ have to be to make the equation true for every possible value of $x$.