How do you solve $T^{2} + 7 T - 2 = 0$ $T^2 + 7T - 2 = 0$ by completing the square?

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Answer 1 · 2016-04-04T10:35:33

$T = - \frac{7}{2} + \frac{\sqrt{57}}{2}$ $T=-7/2+sqrt57/2$ or $T = - \frac{7}{2} - \frac{\sqrt{57}}{2}$ $T=-7/2-sqrt57/2$

As ${(x + a)}^{2} = x^{2} + 2 a x + a^{2}$ $(x+a)^2=x^2+2ax+a^2$ , to complete let us say

$x^{2} + 2 a x$ $x^2+2ax$ to make complete square, we should add $a^{2}$ $a^2$ or square of half of the coefficient of x.

Hence to solve $T^{2} + 7 T - 2 = 0$ $T^2+7T-2=0$ we should add and subtract ${(\frac{7}{2})}^{2} = \frac{49}{4}$ $(7/2)^2=49/4$ .

Hence, $T^{2} + 7 T - 2 = 0$ $T^2+7T-2=0$ can be written as $T^{2} + 7 T + \frac{49}{4} - \frac{49}{4} - 2 = 0$ $T^2+7T+49/4-49/4-2=0$ or

$T^{2} + 7 T + \frac{49}{4} = \frac{49}{4} + 2 = \frac{57}{4}$ $T^2+7T+49/4=49/4+2=57/4$ or

= ${(T + \frac{7}{2})}^{2} = \frac{57}{4}$ $(T+7/2)^2=57/4$

or $T + \frac{7}{2} = \frac{\sqrt{57}}{2}$ $T+7/2=sqrt57/2$ or $T + \frac{7}{2} = - \frac{\sqrt{57}}{2}$ $T+7/2=-sqrt57/2$

or $T = - \frac{7}{2} + \frac{\sqrt{57}}{2}$ $T=-7/2+sqrt57/2$ or $T = - \frac{7}{2} - \frac{\sqrt{57}}{2}$ $T=-7/2-sqrt57/2$

How do you solve T2+7T−2=0T^2 + 7T - 2 = 0 by completing the square?