Question

Solving quadratic equation using the completing method.(a).x^2+6x+1=0​

Answer 1

Answer:

$x = -3 + 2\sqrt2 \ , \ x = -3 - 2\sqrt2$

Step-by-step explanation:

x² + 6x + 1 = 0

a = 1 , b = 6 , c = 1

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\\x = \frac{-6 \pm \sqrt{36-4}}{2}\\\\x=\frac{-6 \pm \sqrt{32}}{2}\\\\x= \frac{-6 \pm \sqrt{16 \times 2}}{2}\\\\x = \frac{-6 +4\sqrt2 }{2}\\\\x = -3 + 2\sqrt2 \ , \ x = -3 - 2\sqrt2$

Answer 2

Answer:

see below

Step-by-step explanation:

Given quadratic:

x² + 6x + 1 = 0

Solution

Take the first two terms, and make a complete square.

If the coefficient of x² is not equal to 1, divide the equation by the coefficient to make the job easier.  It equals one in this case.

We know from

(x+a)² = x² + 2a + a²

so here a =3, or (x+3)² = x² + 6x + 9

Put that in the equation

x² + 6x + 9 -9 +1 = 0

the term -9 is to balance the +9 that we added so as not to change the equation.

Rewrite and simplify

x² + 6x + 9 = 8

(x+3)² = 2(2²)

take square-root on both sides

(x+3) = +/- 2sqrt(2)

therefore

x = -3 +/- 2sqrt(2)

or

x = 3+2sqrt(2),  3-2sqrt(2)