A)square root of 81
B)+-9
C)+- square root of 9
D)can't solve
2007-05-20 16:51:34 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
b) + & - 9 . you meant ''root'' or solution of the quadratic equation. pop these values inthe quadratic equation and see if they satisfy it
2007-05-20 16:54:39 · answer #1 · answered by Dr. Eddie 6 · 0⤊ 0⤋
y^2 - 81 is what we, in the math bidness, call a "special product"
it is in this form:
(a^2 - b^2) or, the difference of squares.
This type of beastie is always equal to:
(a+b)(a-b)
So, in this case, with y^2-81, we're looking at:
(y+9)(y-9)
This thing is going to be equal to zero whenever either y+9 or y-9 is zero.
Which is if y=-9 (the first one) or y=9 (the second.)
y=+-9 are the roots of the quadratic.
2007-05-20 16:55:56 · answer #2 · answered by Roland A 3 · 0⤊ 0⤋
Since you said 'quadratic' I'm guessing that you just can't type and you really meant
y² - 81=0
Then y = ±9
Doug
2007-05-20 16:57:41 · answer #3 · answered by doug_donaghue 7 · 0⤊ 0⤋
i`m not understanding the equation. Is it y to the negative 81st power? And are you supposed to be using the quadratic formula for this problem?
2007-05-20 16:58:58 · answer #4 · answered by shakeybabyy 2 · 0⤊ 0⤋
y^-81=0
That would mean that:
1/(y^81)=0
Which means:
1 = 0
Why is an impossibilty.
Thus: D (Can't Solve)
2007-05-20 16:54:56 · answer #5 · answered by Mr. Good Answers 2 · 0⤊ 0⤋
Y² = 81
Y = ± 9
ANSWER B)
2007-05-20 22:16:52 · answer #6 · answered by Como 7 · 0⤊ 0⤋