2007-09-06 02:41:37 · 5 answers · asked by Michael Ray G 1 in Science & Mathematics ➔ Mathematics
People have provided the answer. I simply want to add some information.
[Remember, (√a)(√a) = a & (√a)(- √a) = - a.]
When you're given the solutions & asked to provide the original equation, you work backward.
Given solutions of √2 & - √2, means:
x = √2
or
x - √2 = 0.
So, (x - √2) is one factor.
&
x = - √2
x + √2 = 0
The second factor is, (x + √2).
The original equation is:
(x - √2)(x + √2) = 0.
Finally, use FOIL to recover the original equation.
2007-09-06 03:17:22 · answer #1 · answered by Anonymous · 0⤊ 0⤋
x^2 - 2 = 0;
2007-09-06 09:46:18 · answer #2 · answered by gjmb1960 7 · 1⤊ 0⤋
(x - â 2)(x + â 2) = 0
x^2 - (â 2)^2 = 0 as a^2 - b^2 = (a+b)(a-b)
u know (â 2)^2 = 2
therefore, x^2 - 2 = 0
2007-09-06 09:48:46 · answer #3 · answered by Anonymous · 1⤊ 0⤋
(x+(square of 2))* (x-(square of 2))
then FOIL it out and you will have a quadratic : )
2007-09-06 09:48:20 · answer #4 · answered by Bear 2 · 1⤊ 0⤋
x^2 -- (sum of roots)x +(product of roots)=0
or x^2 -- ( rt2 -- rt2)x +(rt2)(--rt2) = 0
or x^2 -- 2 = 0
2007-09-06 10:16:03 · answer #5 · answered by sv 7 · 0⤊ 0⤋