Asked by jessica
having a hard time with a couple of problems please help
(3x+5)^2+8=0
also
6-9=12 divided by y+5 y-5 y^2-25
(3x+5)^2+8=0
also
6-9=12 divided by y+5 y-5 y^2-25
Answers
Answered by
bobpursley
On the first, two ways...
Let me do the easiest. Notice that the first term for real x will always be positive, and the second term is positive, the sum is zero. So there are no real x.
(3x+5)^2=-8
take the square root of each side..
3x+5=sqrt (-8)= 2i sqrt2
then solve for x. It will be a complex number.
Second
Multiply both sides by (y^2-25)
6(y-5)-9(y+5)=12
multiply out, gather terms and solve for y.
Let me do the easiest. Notice that the first term for real x will always be positive, and the second term is positive, the sum is zero. So there are no real x.
(3x+5)^2=-8
take the square root of each side..
3x+5=sqrt (-8)= 2i sqrt2
then solve for x. It will be a complex number.
Second
Multiply both sides by (y^2-25)
6(y-5)-9(y+5)=12
multiply out, gather terms and solve for y.
Answered by
jessica
Thank you very much
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