sole the simultaneous equations y=x-4 2x^2-xy=8 giving your answers in the form a plus or minus b root3 , where a and b are integers
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Hello Substitute to eleminate y. 2x^2 -x(x - 4) = 8 Now solve for x.
Hello, In the second equation, replace y by x-4 -> 2x²-x(x-4)=8 --> 2x²-x²+4x=8 ---> x²+4x-8=0 Calculate the discriminant, which is 4²+4*8 = 48 = 16*3 = (4 root 3)² And then, the solutions...
Originally Posted by Moo Hello, In the second equation, replace y by x-4 -> 2x²-x(x-4)=8 --> 2x²-x²+4x=8 ---> x²+4x-8=0 Calculate the discriminant, which is 4²+4*8 = 48 = 16*3 = (4 root 3)² And then, the solutions... i dont know how to solve x²+4x-8=0
One way, complete the square. 2x^2 - x^2 + 4x = 8 x^2 + 4x = 8 (x + 2)^2 - 4 = 8 (x + 2)^2 = 12 x + 2 = +/- root 12 x = -2 +/- 2 root 3
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