Find the real solutions of the equation? (4x-2)^2-8(4x-2)+7=0

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degeneratecircle | High School Teacher | (Level 2) Associate Educator

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This is a quadratic equation in the quantity `4x-2.` If we let `u=4x-2,` our equation becomes `u^2-8u+7=0.` We can factor this by finding two numbers that multiply to `7` and add to `-8,` and a little thought shows that `-7` and `-1` satisfy these requirements. Thus,

`u^2-8u+7=(u-7)(u-1),` and for this to equal zero, we must have

`u=7` or `u=1.` But we want the solutions in terms of `x,` so we go back to the equation `u=4x-2` that relates `u` to `x.` If `u=7,` then `7=4x-2` and `x=9/4.` If `u=1,` then `1=4x-2` and `x=3/4.`

The two solutions are `x=9/4` and `x=3/4.` This can also be done by multiplying everything out and getting `16x^2-48x+27=0` and then factoring or using the quadratic formula, but that is probably the longer way.