What are the solutions for a^2 + 16a = 17?
A) -1 and -7
B) 1 and 17
C) 17 and -1
D) 1 and -17
2007-06-01 02:15:38 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
a^2+16a-17=0
a=((-16+-sqrt(256+68))/2= (-16+-18)/2
x=-17 and x= 1 D)
2007-06-01 02:25:06 · answer #1 · answered by santmann2002 7 · 0⤊ 0⤋
a^2 - 16a - 17 = 0
Seventeen is a prime number, so the only possible way to have a real root is 17 and 1. One of then has to be negative, or 17 would be positive.
-17 + 1 or 17 -1.
The middle term is 16, so use 17-1.
If this didn't work, you'd have to use the quadratic equation.
Anyway
(a - 17)(a + 1) = 0
So a = 17 or a = -1
2007-06-01 09:27:39 · answer #2 · answered by TychaBrahe 7 · 0⤊ 0⤋
a^2 +16a =17
<=> a^2 +16a -17=0
<=>a^2 +17a -a -17=0
<=>a(a+17) - (a+17)=0
<=>(a+17)(a - 1) =0
<=>a +17 =0 and a -1 =0
<=> a= 0 -17 and a= 0 + 1
<=> a= -17 and a= 1
So D is the correct answer.
2007-06-01 09:37:55 · answer #3 · answered by Oriole 1 · 0⤊ 0⤋
a^2 + 16a - 17 = 0
(a - 1) (a +17) = =
a-1=0 so a = 1
and
a+17=0 so a = -17
Remember when you factor.... the numbers multiply to the last number, add to the middle number, 17*-1 = -17 and 17+-1=16
2007-06-01 09:30:29 · answer #4 · answered by Anonymous · 0⤊ 0⤋
a^2+16a-17=0
(a-1)(a+17)=0
therefore solutions = 1 and -17
D
2007-06-01 09:26:51 · answer #5 · answered by Anonymous · 0⤊ 0⤋
well it can definitely be 1 :P
(1*1)+16=17
don't have calculator with me.. anyone know a free graphical calculator program?
2007-06-01 09:26:36 · answer #6 · answered by bob e 1 · 0⤊ 0⤋