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### Find the possible values of x for x^2 = 36-5x.

While this question may appear unfamiliar at first, it is just a quadratic equation in disguise. Rearranging such that it equals 0:

x^2 + 5x - 36 = 0

Now this equation is ready to be solved! We can approach this in a couple of ways. I'll go through the factorisation method and using the quadratic formula.

Factorisation Method

This method is often the quickest way but requires a bit of practise to be able to spot the solution. If we look at a general factorisation where the coefficient of x is 1 in both brackets, i.e.

(x+a)(x+b) = 0

We can see when we expand this, we get

x^2 + ax + bx + ab = 0

x^2 + (a+b)x + ab = 0

Going back to our example, this shows us that we need to think of numbers such that a + b = 5 and ab = -36. The tricky part is actually thinking of these numbers.

Because we know that ab is a negative number, we know either a or b must be negative because a negative times a positive is a negative. We now need to trial and error a few potential numbers to find our answers. With a bit of good guessing we can see that a = -4 and b = 9 will work because -4x9 = -36 and -4+9 = 5.

This gives us

(x-4)(x+9)=0

For this to equal zero, one of these brackets must equal 0. If the first one equals 0 then,

x-4 = 0

x=4

or the second one,

x+9=0

x=-9

So our answer is x=4 or x=-9.

If you found it difficult to guess the right numbers then keep practising but, in the meantime, the formula method always works without guessing but does take a little longer...

Formula Method

For a general quadration equation ax^2 + bx + c = 0, the quadratic formula is given by

x = (-b +/- sqrt(b^2 - 4ac))/2a

For our example, a = 1, b = 5 and c = -36, so:

x = (-5 +/- sqrt(25 + (4x36)))/2

x = (-5 +/- sqrt(25+144))/2

x = (-5 +/- sqrt(169))/2

x = (-5 +/- 13)/2

Taking the + option:

x = (-5 + 13)/2

x = 8/2

x = 4

Taking the - option:

x = (-5-13)/2

x = -18/2

x = -9

Therefore, x=4 or x=-9 as expected from our first method.

11 months ago

Answered by Ben, a GCSE Maths tutor with MyTutor

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