# How do you solve by completing the square a^2 -6a-5=0?

Jun 8, 2018

Solution: $a = 3 + \sqrt{14} \mathmr{and} a = 3 - \sqrt{14}$

#### Explanation:

${a}^{2} - 6 a - 5 = 0 \mathmr{and} {a}^{2} - 6 a = 5$ or

${a}^{2} - 6 a + 9 = 5 + 9$ or

${\left(a - 3\right)}^{2} = 14 \mathmr{and} a - 3 = \pm \sqrt{14}$ or

$a = 3 \pm \sqrt{14}$

Solution: $a = 3 + \sqrt{14} \mathmr{and} a = 3 - \sqrt{14}$ [Ans]

Jun 8, 2018

$a = 3 + \sqrt{14} , a = 3 - \sqrt{14}$

#### Explanation:

When we want to make a square out of a quadratic polynomial, we need to look at the b-coefficient of the polynomial (assuming $0 = a {x}^{2} + b x + c$).

We know that ${\left(a - b\right)}^{2} = {a}^{2} - 2 a b - {b}^{2}$.
Therefore our "a" is just going to be a, and b is going to be some number for which :

$- 2 a b = - 6 a$
$- 2 b = - 6$
$b = 3$

So our square is going to be ${\left(a - 3\right)}^{2}$. However, the square equals to ${a}^{2} - 6 a + 9$ and we lack the +9 to complete the square.
Luckily, we are free to add a zero to the equation anytime we want. Like this :

${a}^{2} - 6 a + \left(9 - 9\right) - 5 = 0$
${a}^{2} - 6 a + 9 - 9 - 5 = 0$
${\left(a - 3\right)}^{2} - 14 = 0$
${\left(a - 3\right)}^{2} = 14$

Now we can square the equation, watch out for the plus/minus signs :

$| a - 3 | = \sqrt{14}$
$a - 3 = \pm \sqrt{14}$
$a = 3 \pm \sqrt{14}$

And we are finished.