x^2 + 8x – 16
2006-12-06 04:17:27 · 8 answers · asked by Deidrebm 1 in Science & Mathematics ➔ Mathematics
Check out the discriminant b²-4ac. a=1, b=8, c=-16.
b²-4ac = 64 - 4*1*(-16) = 128 = not zero, so it isn't a square.
2006-12-06 04:22:38 · answer #1 · answered by Anonymous · 1⤊ 0⤋
No.
An alternate way to do it is that if it were a perfect square, i.e. x^2 + 8x - 16 = (ax + b)^2, by virtue of the squaring, it would take only nonnegative values. In particular, if we plug in x=0, then the value of the polynomial is -16, so this would mean (a*0 + b)^2 = -16, which doesn't work when we work in real numbers. Cheers!
2006-12-06 13:09:05 · answer #2 · answered by bag o' hot air 2 · 0⤊ 0⤋
No its not, it would have to be x^2+8x+16 which would become (x+4)^2 or it can be x^2-8x+16 which would be come (x-4)^2
2006-12-06 12:24:40 · answer #3 · answered by gelal2003 2 · 1⤊ 0⤋
No.
you should have +16
2006-12-06 12:34:00 · answer #4 · answered by maussy 7 · 0⤊ 0⤋
A perfect square is an algebraic expression that can be factored as the square of some other expression.
Use the quadratic formula to factor your expression. You will find that the factors are not identical, and so your expression is not a perfect square.
2006-12-06 12:39:05 · answer #5 · answered by Jerry P 6 · 0⤊ 0⤋
x^2+8x-16
(x+4)(x-4)
i say it isn't a perfect square, but...
x^2+8x+16 is a perfect square, since you would end up with
(x+4)(x+4) = (x+4)^2
now stop letting others do your homework! lol
2006-12-06 12:24:27 · answer #6 · answered by cadpro78 2 · 0⤊ 1⤋
if the signs of x^2 and the constant term are different, it can never be a perfect square, or negative of one.
2006-12-06 12:29:10 · answer #7 · answered by astrokid 4 · 3⤊ 0⤋
ya...sure!!!
2006-12-06 12:23:22 · answer #8 · answered by Choclate thunder 2 · 0⤊ 2⤋