How do you solve using the completing the square method $2 x^{2} - 8 x - 3 = 0$ $2x^2-8x-3=0$ ?

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Answer 1 · 2018-06-15T20:18:54

See below

First extracting common factor

$2 (x^{2} - 4 x - \frac{3}{2}) = 0$ $2(x^2-4x-3/2)=0$

Now complete square

$2 ({(x - 2)}^{2} - 4 - \frac{3}{2}) = 0$ $2((x-2)^2-4-3/2)=0$ now re-arrange terms

$2 ({(x - 2)}^{2} - \frac{11}{2}) = 0$ $2((x-2)^2-11/2)=0$

Now, using identity $(a + b) (a - b) = a^{2} - b^{2}$ $(a+b)(a-b)=a^2-b^2$

$2 ((x - 2 + \sqrt{\frac{11}{2}}) (x - 2 - \sqrt{\frac{11}{2}})) = 0$ $2((x-2+sqrt(11/2))(x-2-sqrt(11/2)))=0$

Roots are $2 + \sqrt{\frac{11}{2}}$ $2+sqrt(11/2)$ and $2 - \sqrt{\frac{11}{2}}$ $2-sqrt(11/2)$

How do you solve using the completing the square method 2x2−8x−3=02x^2-8x-3=0?