How do you complete the square to solve $4x^2 - 7x - 2 = 0$ ?

Question

7211 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-23T14:37:38

In this way:

$4x^2-7x-2=0rArr4(x^2-7/4x)-2=0rArr$

$4(x^2-7/4x+49/64-49/64)-2=0rArr$

$4(x^2-7/4x+49/64)-49/16-2=0rArr$

$4(x-7/8)^2=49/16+2rArr4(x-7/8)^2=(49+32)/16rArr$

$(x-7/8)^2=81/16*1/4rArr(x-7/8)^2=81/64rArr$

$x-7/8=+-9/8rArrx=7/8+-9/8rArr$

$x_1=7/8-9/8=-2/8=-1/4$

$x_2=7/8+9/8=16/8=2$ .

How do you complete the square to solve 4x^2 - 7x - 2 = 04x^2 - 7x - 2 = 0?