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We have to prove: (2x-1)^2-(x-3)^2 = (3x-4)(x+2)
Take the left hand side
(2x-1)^2-(x-3)^2
Use the relation x^2 - y^2 = (x - y)(x + y)
=> (2x - 1 - x + 3)(2x - 1 + x - 3)
=> (x + 2)(3x - 4)
which is the right hand side.
This proves that (2x-1)^2-(x-3)^2 = (3x-4)(x+2)
We'll raise to square using the formula:
(a+b)^2 = a^2 + 2ab + b^2
(a-b)^2 = a^2 - 2ab + b^2
(2x-1)^2 = (2x)^2 - 2*(2x)*1 + 1^2
(2x-1)^2 = 4x^2 - 4x + 1 (1)
(x-3)^2 = x^2 - 2*x*3 + 3^2
(x-3)^2 = x^2 - 6x + 9 (2)
If we'll subtract (2) from (1), we'll get:
4x^2 - 4x + 1 - x^2 + 6x - 9
We'll combine like terms:
3x^2 + 2x - 8
We'll cancel the quadratic to determine it's x intercepts:
3x^2 + 2x - 8 = 0
x1 = [-2+sqrt(4+96)]/6
x1 = (-2+10)/6
x1 = 8/6
x1 = 4/3
x2 = (-2-10)/6
x2 = -2
We'll write the quadratic as a product of linear factors:
3x^2 + 2x - 8 = 3(x - 4/3)(x + 2)
We'll multiply by 3 the first factor:
3(x - 4/3)(x + 2) = (3x - 4*3/3)(x + 2)
3x^2 + 2x - 8 = (3x-4)(x+2)
(2x-1)^2 - (x-3)^2 = (3x-4)(x+2)
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