Solving a quadratic with ax^2 e.g. 2x^2 - 11x + 12 = 0
To solve the equation in the form (ax + b)(x + c):
Method 1
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Rewrite the x term as the sum of 2 different x terms. Start by multiplying the x^2 term and the final number term, to give 24. The two x terms should add to give the middle term, and multiply to give 24: 2x^2 - 11x + 12 = 2x^2 - 8x -3x +12 >>> the first "half" of the equation, 2x^2 - 8x, has common factors with the second "half", -3x +12 and the sum of -8 and -3 is 24, while they add to -11.
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Factorise this expression - always ensure that the terms in the brackets match 2x^2 - 8x - 3x + 12 = 2x(x-4) -3(x-4) 3. Both the 2x and the -3 have a common factor - factorise further 2x(x-4) -3(x-4) = (x-4)(2x-3)
Method 2 Start by creating a pair of brackets - to get the 2x^2 term, one bracket must contain x and the other must contain 2x: (x + a)(2x + b) Looking at the final term, 12 - this can be obtained in several ways e.g -6 & -2, 3 & 4 so the right pair of numbers can be found by looking at the middle term, -11x. If we multiply out our brackets, we get 2x^2 + bx +2ax + ab. We can use simultaneous equations to find a and b as we know that ab = 12 and bx + 2ax = - 11x, or b + 2a = -11. This could be solved algebraically, but at this level the standard quadratic will contain integers. We can try pairs of numbers (3&4, 6&2, 12&1 both + and -) to find which pair fits the second equation, b + 2a = -11. The only pair that fits this term is b = -3 and a = -4. Therefore our quadratic is (x-4)(2x-3)
