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

Question

Genesis Allen · Accepted Answer

#T=-7/2+sqrt57/2# or #T=-7/2-sqrt57/2#

Since #(x+a)^2=x^2+2ax+a^2#, let's finish by saying

To get the complete square, we need to add #a^2#, which is the square of half of the coefficient of x. This is #x^2+2ax#.

Therefore, we must add and subtract #(7/2)^2=49/4# in order to solve #T^2+7T-2=0#.

Therefore, it is possible to write #T^2+7T-2=0# as #T^2+7T+49/4-49/4-2=0# or

T^2 + 7T + 49/4 = 49/4 + 2 = 57/4# or

= #(T+7/2)^2=57/4#

Alternatively, #T+7/2=-sqrt57/2# or #T+7/2=sqrt57/2#

alternatively #T=-7/2+sqrt57/2# or #T=-7/2-sqrt57/2#

Avery Alexander · Answer

undefined To solve the quadratic equation $ T^2 + 7T - 2 = 0 $ by completing the square, follow these steps:

1. Move the constant term to the other side of the equation:
   $ T^2 + 7T = 2 $

2. Add and subtract half of the coefficient of $ T $ squared (in this case, $ \frac{7}{2} $ squared) to both sides of the equation:
   $ T^2 + 7T + \left(\frac{7}{2}\right)^2 = 2 + \left(\frac{7}{2}\right)^2 $

3. Simplify both sides of the equation:
   $ T^2 + 7T + \frac{49}{4} = 2 + \frac{49}{4} $

4. Factor the left side of the equation as a perfect square:
   $ \left(T + \frac{7}{2}\right)^2 = \frac{57}{4} $

5. Take the square root of both sides of the equation:
   $ T + \frac{7}{2} = \pm \sqrt{\frac{57}{4}} $

6. Solve for $ T $:
   $ T = -\frac{7}{2} \pm \sqrt{\frac{57}{4}} $