Question

PLS HELP: How do you solve x^2-ax-6a^2=0 by completing the square?

Answer 1

Answer:

(-3a+-2x)(2a+-1x)=0

Step-by-step explanation:

Simplifying

x2 + -1ax + -6a2 = 0

Reorder the terms:

-1ax + -6a2 + x2 = 0

Solving

-1ax + -6a2 + x2 = 0

Solving for variable 'a'.

Factor a trinomial.

(-2a + -1x)(3a + -1x) = 0

Answer 2

Answer:

x = a/2 + 6.5 a^2

Step-by-step explanation:

x^2-ax-6a^2=0

Assuming a is a constant, let's move the third term to the right:

x^2-ax=6a^2

Now we need to add something to both sides to make it a perfect square.

x^2-ax + _____ = 6a^2 + ______

We need to add a/2 ^2 to both sides. (This is a standard part of completing the square, wherein you add b/2 ^2. Like if you had 8x as your middle term, you add 16.)

x^2-ax + a/2 ^2 = 6a^2 + a/2 ^2

Factor the left side, because it's a perfect square now:

(x - a/2)(x - a/2) = 6.5 a^2

(x - a/2)^2 = 6.5 a^2

Take the square root of both sides:

x - a/2 = 6.5 a^2

x = a/2 + 6.5 a^2