SOLUTION: I can't seem to get these right - help would be appreciated: (Not necessary to find the roots - just determine the number & type of solution) use a discriminant to determine the

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I can't seem to get these right - help would be appreciated: (Not necessary to find the roots - just determine the number & type of solution) use a discriminant to determine the      Log On


   



Question 59581: I can't seem to get these right - help would be appreciated:
(Not necessary to find the roots - just determine the number & type of solution) use a discriminant to determine the number of solutions for the quadratic equasions - and if the solutions are real or complex.
a. 2x^2+x-1=0
b. 4/3x^2-2x+3/4=0
c. m^2+m+1=0
Thank you very much!

Found 3 solutions by stanbon, uma, funmath:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
a. 2x^2+x-1=0
discrim=1^2-4*2*-1=9; 2 real zeroes
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b. 4/3x^2-2x+3/4=0
multiply thru by 12 to get:
16x^2-24x+9=0
discrim= 24^2-4*16*9=0; 2 identical real zeroes
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c. m^2+m+1=0
discrim=1^2-4*1*1=-3 ; 2 complex zeroes.
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Cheers,
Stan H.

Answer by uma(370) About Me  (Show Source):
You can put this solution on YOUR website!
1. 2x^2 + x - 1 = 0
Here a = 2, b = 1 and c = -1
Discriminant D = b^2 - 4ac
= 1^2 - 4(2)(-1)
= 1 +8
= 9
9 is positive and a perfect square.
So the two roots of the given equation are real and rational.

b) 4/3 x^2 -2x + 3/4 = 0
Here a = 4/3, b= -2 and c= 3/4
Discriminant = D = b^2 - 4ac
= (-2)^2 - 4(4/3)(3/4)
= 4 - 4
= 0
As the discriminant is zero, the given equation has two real and equal roots.
c) m^2 + m + 1 = 0
Here a = 1, b = 1 and c = 1.
Discriminant = D = b^2 - 4ac
= 1^2 - 4(1)(1)
= 1 - 4
= -3
D is negative and so both the roots of the given equation are complex.
Good Luck!!!





Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
(Not necessary to find the roots - just determine the number & type of solution) use a discriminant to determine the number of solutions for the quadratic equasions - and if the solutions are real or complex.
:
The discriminant of a quadratic equation written in standard form ax%5E2%2Bbx%2Bc=0 is highlight%28b%5E2-4ac%29
If the discriminant is positive, there are two real solutions.
If the discriminant is 0, there is one real solution.
If the discriminant is negative, there are two complex solutions.
:
a. 2x^2+x-1=0 a=2, b=1, c=-1
b%5E2-4ac=%281%29%5E2-4%282%29%28-1%29
=1%2B8=9 The discriminant is positive, there are two real solutions.
:
b. 4/3x^2-2x+3/4=0 a=4/3, b=-2, c=3/4
b%5E2-4ac=%28-2%29%5E2-4%284%2F3%29%283%2F4%29
=4-4%2812%2F12%29%29
4-4%281%29
4-4
0 The discriminant is 0, there is one real solution.
:
c. m^2+m+1=0 a=1, b=1, c=1
b%5E2-4ac=%281%29%5E2-4%281%29%281%29
=1-4
=-3 The discriminant is negative, there are two complex solutions.
:
Happy Calculating!!!