can someone please show me how to solve the quadratic equation x^2-11=2x?

Question

I have gotten this far:
x^2-2x-11=0
-b+or-sq. rt. of b^2-4ac/2a
2+or- sq.rt. of (-2)^2-4(1)(-11)/2
2+or- sq.rt. of 48/2

I can't figure out what to do to simplify the equation after. Since the sq. rt. of 48 is not a real #, what do we do to simplify?

Answer 1

The square root of 48 is a real number, it just isn't a rational number. You simplify sqrt(48) like so:

sqrt(48)
= sqrt(16 * 3)
= sqrt(16) * sqrt(3)
= 4 * sqrt(3)

So your answer is then
x = [ 2 +/- sqrt(48) ] / 2
x = [ 2 +/- 4 * sqrt(3) ] / 2
x = 1 +/- 2 * sqrt(3)

Hope this helps

Answer 2

x² - 11 = 2x
x² - 2x - 11 = 0
a = 1, b = -2, c = -11, so

x = -(-2) / 2(1) ± √[ (-2)² - 4(1)(-11)] / 2(1)
x = 1 ± √(4 + 44) / 2
x = 1 ± √48 / 2
x = 1 ± √[16(3)] / 2
x = 1 ± 4√3 / 2
x = 1 ± 2√3

notice it's almost more work to simplify the answer than to fill in the quadratic formula.

completing the square might be simpler in this case:
x² - 2x - 11 = 0
x² - 2x + 1 = 11 + 1
(x - 1)² = 12
x - 1 = ±√12 = ± 2√3
x = 1 ± 2√3

Answer 3

The answer lies in making the factors of 48 such that one of it is a perfect square. The factors of 48 are 2, 3, 4, 6, 8, 12, 16, 24, 48.

Now, you see that 4 and 16 are the perfect squares in the factors of 48. We pick 16 because that would give the most simpified form.
We can write sqrt(48) as sqrt(16 * 3).
This can be written as sqrt(16) * sqrt(3) = 4 sqrt(3)

Therefore, your x = 1 +- 2 sqrt(3)

Answer 4

Always, always put it into the usual form:
ax^2 + bx + c = 0
x^2 - 2x -11 = 0, so a = 1, b = -2, c = -11
x = {-(-2) +- sqr[(-2)^2 - 4(1)(-11)]}/2(1)
= {2 +- sqr[4 +44]}/2 = {2 +- sqr[48]}/2
= {2 +- sqr[16 x 3]}/2 = {2 +- 4sqr[3]}/2
= 1 +- 2sqr[3]

Answer 5

i will continue from where u have reached
till there it is correct.
(2 +/- sqrt(48))/2
(2 +/- sqrt(16*3))/2
(2 +/- sqrt(16)*sqrt(3))/2
(2 +/- 4*sqrt(3))/2
(2 +/- 4*1.73)/2
(2 +/- 6.92)/2
8.92/2 or -4.92/2
4.46 or -2.46


thus the ans is 4.46 or -2.46