SOLUTION: determine whether the equation has 2 solutions, 1 solution, or no solution. SHOW WORK! PUT IN STANDARD FORM! 22)x^2-3x+2=0 24)-3x^2+5x-1=0 26)x^2-2x+4=0 28)3x^2-6x+3=0 30

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: determine whether the equation has 2 solutions, 1 solution, or no solution. SHOW WORK! PUT IN STANDARD FORM! 22)x^2-3x+2=0 24)-3x^2+5x-1=0 26)x^2-2x+4=0 28)3x^2-6x+3=0 30      Log On


   

Question 119529This question is from textbook
: determine whether the equation has 2 solutions, 1 solution, or no solution. SHOW WORK! PUT IN STANDARD FORM!
22)x^2-3x+2=0
24)-3x^2+5x-1=0
26)x^2-2x+4=0
28)3x^2-6x+3=0
30)-5x^2+6x-6=0 This question is from textbook

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do the first two to help you get started


#22

From the quadratic formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

the discriminant consists of all of the terms in the square root. So the discriminant is

D=b%5E2-4ac

the discriminant tells us how many solutions (and what type of solutions) we can expect for any quadratic.


Now let's find the discriminant for y=x%5E2-3x%2B2:

D=-3%5E2-4%2A1%2A2 Plug in a=1, b=-3, c=2

D=9-4%2A1%2A2 Square -3 to get 9

D=9-8 Multiply -4*1*2 to get -8

D=1 Combine 9 and -8 to get 1


Since the discriminant equals 1 (which is greater than zero) , this means there are two real solutions. Remember if the discriminant is greater than zero, then the quadratic will have two real solutions.







#24


From the quadratic formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

the discriminant consists of all of the terms in the square root. So the discriminant is

D=b%5E2-4ac

the discriminant tells us how many solutions (and what type of solutions) we can expect for any quadratic.


Now let's find the discriminant for y=-3x%5E2%2B5x-1:

D=5%5E2-4%2A-3%2A-1 Plug in a=-3, b=5, c=-1

D=25-4%2A-3%2A-1 Square 5 to get 25

D=25-12 Multiply -4*-3*-1 to get -12

D=13 Combine 25 and -12 to get 13


Since the discriminant equals 13 (which is greater than zero) , this means there are two real solutions. Remember if the discriminant is greater than zero, then the quadratic will have two real solutions.