# Rationalize denominator & factorising quadratic equations.

#### 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.

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#### andrewkirk

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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.
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: $$\frac{a}{b}\times\frac{c}{d}=\frac{a\times c}{b\times d}$$.

#### andrewkirk

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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.
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 $x$.

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