(2 ± √-56)/6 is right so far
now √-56 = (√-1)(√4)√14) = (2√14)i
so
(2 ± √-56)/6
= (2 ± (2√14)i)/6
= (1 ± √14 i)/3
Can you explain how to find the solution of 3x^2-2x+5, using the quadratic formula? I got it down to (2 +/- sq. root of -56)/6. But my book then reduces that to (2 +/- 2 times the sq. root of 14i)/6. But I think in place of the 2 and the sq. root of 14 it could be the sq. root of 7i and the sq. root of four and two, which equals eight. This would then break down to 2 the sq. root of 7i sq. root of 4. But how come this isn't an answer? Why is the book's answer right, and how do you know to choose that one over others? Thanks!!
3 answers
How come 14 doesn't get reduced to the sq. root of 7i. I would think you'd divide by two on the top and bottom: 6/2=3, 14/2=7).
I don't know that but mine work Is on slope and I don't know it so can you please help :)
1.(1,5)(2,7) 2.(0,1)(3,-8) 3.(2,-3)(4,-2) 4.(2,5)(4,2) 5.(-3,-5)(-1,3) 6.(3,-1)(-6,-4) 7.(4,1)(-4,7) 8.(-1,2)(3,4) 9.(-1,-4)(2,0) 10.(3,-1)(-3,5)
1.(1,5)(2,7) 2.(0,1)(3,-8) 3.(2,-3)(4,-2) 4.(2,5)(4,2) 5.(-3,-5)(-1,3) 6.(3,-1)(-6,-4) 7.(4,1)(-4,7) 8.(-1,2)(3,4) 9.(-1,-4)(2,0) 10.(3,-1)(-3,5)
1 answer