How do you solve $2 x^{2} - 4 x - 3 = 0$ $2x^2-4x-3=0$ by completing the square?

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Answer 1 · 2016-07-11T13:36:14

$x = 1 \pm \sqrt{\frac{5}{2}} .$ $x=1+-sqrt(5/2).$

Taking, $\sqrt{\frac{5}{2}} ≅ 1.581$ $sqrt(5/2)~=1.581$ , we get the soln. set $= {2.581, - 0.581} .$ $={2.581, -0.581}.$

Given that, $2 x^{2} - 4 x - 3 = 0 .$ $2x^2-4x-3=0.$

$:. 2x^2-4x=3.$

To complete the square on $L.H.S.$ , we need
Last Term $= (Midl.Term)^2/(4xx First Term)=(-4x)^2/(4*2x^2)=(16x^2)/(8x^2)=2$

Adding $2$ on both sides,

$2x^2-4x+2=3+2$

$:. 2(x^2-2x+1)=5$

$:. 2(x-1)^2=5$

$:. (x-1)^2=5/2$

$:. (x-1)=+-sqrt(5/2)$

$:. x=1+-sqrt(5/2),$ is the desired soln.

Taking, $sqrt(5/2)~=1.581$ , we get the soln. set $={2.581, -0.581}.$

How do you solve 2x^2-4x-3=02x2−4x−3=02x^2-4x-3=0 by completing the square?