How do you solve the quadratic with complex numbers given #10x^2-11x+9=13x-6x^2#?

Question

3236 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-03T02:50:50

I think you want to solve for x, right?

#x = 3/4#

There is only one solution.

#6x^2+10x^2-11x-13x+9=0#

#16x^2-24x+9=0#

#x=(-(-24)+-sqrt((-24)^2-4(9*16))) / (2*16)#

#x=(24+-sqrt((576-4(144))))/(32)#

#x=(24+-sqrt((576-576)))/(32)#

#x = 24/32 = 3/4#

Since under the square root is equal to 0, this means that there is only one solution.

From looking at the graph, we can also see that it is equal to #3/4#
graph{16x^2-24x+9 [-1.276, 2.436, -0.928, 0.928]}

Answer 2 · 2017-03-06T15:44:36

#x = 3/4#

#10x^2 -11x+9 =13x-6x^2" "larr# make a quadratic equal to 0

#10x^2 +6x^2 -11x -13x +9 =0#

#16x^2 -24x +9=0" "larr# find factors

(Find factors of 16 and 9 which add to give 24.)

#(4x-3)(4x-3)=0#

#4x-3 =0#

#4x = 3#

#x =3/4#

As the two brackets are the same, there is only one solution.