is -2+3x=2x^2
-2x^2+3x-2=0 in standard form? I believe it is but I do quadratic and its no solutions so not X intercept but just want to double check I did the standard form right. Thanks.
2007-08-28 11:19:33 · 4 answers · asked by Alfonso C 1 in Science & Mathematics ➔ Mathematics
Yes,it is in correct form.But in the next step please multiplying all the terms by -1 to get
2x^2-3x+2=0
And then proceed as follows
=>2x^2-2x-x+2=0
=>2x(x-2)-1(x-2)=0
=> (x-2)(2x-1)=0
Therefore either x-2=0 or 2x-1=0
x=2 or 1/2
Hope this helps you
2007-08-28 11:27:48 · answer #1 · answered by Anonymous · 0⤊ 1⤋
2x^2-3x+2=0
x=[3+/- sqrt(9-16)]/4
=[3+/-sqrt(-7)]/4 ANS.
2007-08-28 18:35:28 · answer #2 · answered by Anonymous · 0⤊ 0⤋
-2x^2+3x-2=0
x = [-3 +/- sqrt(9-4(-2)(-2))]/-4
x = [-3 +/-sqrt(-7)]/-4
Thus there are no real roots since x is imaginary
2007-08-28 18:27:54 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
Yep, it's the only solution, you might've accidentaly checker it over wrong, lol that happens to me.
2007-08-28 18:26:35 · answer #4 · answered by Anonymous · 0⤊ 0⤋