Use the quadratic formula to solve each of the following quadratic equations.?

Question

1#
2x^2-5x=3
2#
3x^2-2x+1=0

Answer 1

Ok, I assume you know the quadratic equation, but just for convenience I'm going to write it out: x = (-b +/- root(b^2 - 4ac))/2a.

1) x = (5 +/- root(25 + 24))/4. (Remember, you have to subtract the 3 from each side to get the equation to equal 0, so c is -3.) That simplifies to x = (5 +/- 7)/4--->x = 12/4 or -2/4--->x = 3 or -1/2.

2) x = (2 +/- root(4 - 12))/6. Are you sure that wasn't supposed to be a negative 3x^2 or minus 1? If it were, the term inside the square root would be 16, a positive perfect square. As it is, it's -8, whose square root is 2root(2)i. That seems pretty complex (no pun intended) compared to the other one.

I'm going to solve that one assuming that it should have ended in -1 rather than 1. Then x = (2 +/- root(4 + 12))/6--->x = (2 +/- 4)/6--->x = -2/6 or 6/6--->x = -1/3 or 1.

Answer 2

quadratic formulation is x= -b+root(b^2-4ac) all divided by ability of 2a. and x= -b-root(b^2-4ac) all divided by ability of 2a. u would desire to apply the two equations to get all solutions. a equals the variety in front of the x^2 that's 2. b equals the variety in front of the x that's -5. and c is the consistent whilst on the left area of the equation that's -3. plug the a,b and c into the quadratic formulation and u get: 5+root{(-5)^2 -(4x2x-3)} all divided by ability of four. besides as: 5-root{(-5)^2 -(4x2x-3)} all divided by ability of four. that's 5+(25+24) all divided by ability of four and 5 - (25+24) all divided by ability of four. and u get x=3 and x=-a million/2.

Answer 3

2x^2-5x=3
2x^2-5x-3=0
a=2, b= -5, c=-3
-b + or - [ sqrt (b^2 - 4ac) ] all over 2a

5 +/- sqrt [ 5^2 - (4*2*-3) ] all over 2*2
5 +/- sqrt [ 25 + 24 ] all over 4
5 +/- sqrt [49] all over 4
[ 5 +/- 7 ] / 4
12/4 and -2/4
3 and -1/2 <----answer

3x^2-2x+1=0
a=3, b= -2, c=1
-b + or - [ sqrt (b^2 - 4ac) ] all over 2a
2 +/- [ sqrt ( 4 - 4*3*1 ) all over 2*3
2 +/- [sqrt 4-12] all over 6
2 +/- [sqrt -8] all over 6
[2 +/- 2i*(sqrt 2)] / 6 <----answer or simplify to
[ 1 +/- i(sqrt 2) ] / 3 or (1/3) +/- [ i*(sqrt 2) ] / 3

3x^2-2x-1=0 <-----in case you meant -1
a=3, b= -2, c=-1
-b + or - [ sqrt (b^2 - 4ac) ] all over 2a
2 +/- [ sqrt ( 4 - 4*3*-1 ) all over 2*3
2 +/- [sqrt 16] all over 6
[2 +/- 4] / 6
1 and -(1/3) <----answer

Answer 4

is the statement above a question? if it is, it doesn't make any sense to me

Answer 5

damn, do your own homework