sqrt = squareroot
^2 = to the 2nd power.
Ok heres the problem:
x^2 - 10x + 21 = 0
a=1 b= -10 c= 21 Right?
x= -b +/- sqrt b^2 - 4(a)(c) over 2(a)
Thats the formula
Now does b go in the formula as -10? or 10 because the formula says -b not b so would that make it 10?
2007-11-20 10:21:51 · 4 answers · asked by net_phantom16 2 in Science & Mathematics ➔ Mathematics
I got it already guy, but thanks a lot, heres another question:
If I get a - under the radical what do I do?
x = 3 +/- sqrt -11 over 2
what do I do with the -11 under the radical?
2007-11-20 10:31:55 · update #1
OK
here's the setup
10 +- sqrt (100 - (4)(1)(21)) / 2(1)
10 +- sqrt(100 -84) /2
10+-sqrt (16) / 2
Answers - (10 + 4)/2 and (10-4)/2
= 7 , 3
Hope that helps.
2007-11-20 10:32:25 · answer #1 · answered by pyz01 7 · 0⤊ 0⤋
x^2 - 10x + 21 = 0
a=1 b= -10 c= 21 Right?
x= -b +/- sqrt b^2 - 4(a)(c) over 2(a)
x= -(-10) +/- sqrt (-10)^2 -4(1)(21) / 2(1)
x= 10 +/- sqrt (100-84)/2
etc.
2007-11-20 18:27:02 · answer #2 · answered by ¿ /\/ 馬 ? 7 · 0⤊ 0⤋
-(-10)=10
so it would be 10+-sqrt(100-84)/2=10+-4/2=7 or4
2007-11-20 18:27:35 · answer #3 · answered by someone else 7 · 0⤊ 0⤋
yes.
2007-11-20 18:25:36 · answer #4 · answered by j.investi 5 · 0⤊ 0⤋