2007-05-23 10:50:13 · 6 answers · asked by JN N 1 in Science & Mathematics ➔ Mathematics
solve what? it is not an equation.
do you mean = 0? or maybe =11??
Let me know!
We need an equal sign for it to be an equation.
If you meant 2x^2 + 10x + 11 = 0 then
2x^2 + 10x = -11
x^2 + 5x = -11/2
x^2 + 5x +___= -11/2 + ___
x^2 + 5x +_25/4_ = -11/2 + _25/4_
(x + 5/2)^2 = 3/4
x+ 5/2 = +/-sqrt(3/4)
x = -5/2 +/-sqrt(3/4)
= -5/2 +/- sqrt3 / 2
= [ -5 +/- sqrt3 ] / 2
=]
..
2007-05-23 10:53:32 · answer #1 · answered by Anonymous · 0⤊ 0⤋
I don't know exactly what you mean by the question but i'll give it a try
2x^2+10x+11
= 2(x^2+10x+11)
= 2((x+5)^2)+11-25
*subtract 11-25 then multiply by 2 which is the coeffficiant of x^2
= 2((x+5)^2)-28
2007-05-23 11:45:52 · answer #2 · answered by Lailah 1 · 0⤊ 0⤋
x^2+5x=-11/2
x^2+5x+25/4=-22/4+25/4
(x+5/2)^2=3/4
x+5/2=√3/4
x=-5/2+-√3/4
(the √ is the supposed to be a square root sign)
2007-05-23 10:58:10 · answer #3 · answered by vols_fan 2 · 0⤊ 0⤋
2x^2+10x+11 = 0
x^2 +5x +5.5=0
x^2 +5x + 6.25 +5.5 = 6.25
(x+2.5)^2 = .75
x+2.5 = sqrt(.75) = sqrt(.25 *3)
x+2.5 = .5sqrt(3)
x = .5sqrt(3) -2.5, and
x = - .5sqrt(3)-2.5
2007-05-23 11:03:38 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋
2(x^2+5x+25/4-25/4)+11
2(x+5/2)^2-25/4(2)+11
2(x+5/2)^2-50/2+22/2
2(x+5/2)^2-28/2
2(x+5/2)^2-14=0
Now solve for x.
2007-05-23 10:59:52 · answer #5 · answered by Anonymous · 0⤊ 0⤋
2x^2+10x+11=0 divide by 2
x^2+5x+5.5=0
x^2+5x+2.5^2-2.5^2+5.5=0
(x+2.5)^2-0.75=0
(x+2.5)^2=0.75
x+2.5=±0.5√3
x=-2.5+0.5√3
x=-2.5-0.5√3
2007-05-23 11:08:32 · answer #6 · answered by yupchagee 7 · 0⤊ 0⤋