SOLUTION: X^2-4x+c=0. I have to use discriminant.

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Question 781819: X^2-4x+c=0. I have to use discriminant.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2-4x%2Bc=0...where a=1, b=-4, and c=c
The discriminant D is the part of the quadratic formula underneath the radical sign: the D=b%5E2+-+4ac part
D=b%5E2+-+4ac...plug in a=1, b=-4, and c=c
D=%28-4%29%5E2+-+4%2A1c
D=16+-+4c

If D%3E0, the equation has 2 real solutions.
If D=0, the equation has 1 real solution.
If D%3C0, the equation has 2 conjugate imaginary solutions.
so,
D=16+-+4c...set D less then 0 and find c
0%3C16+-+4c
4c%3C16
c%3C4
if c%3C4 you will have positive discriminant and that means you will have 2 real solutions
D=16+-+4%2A3
D=16+-+12
D=4 => D%3E0
check it: take c=3, plug it in x%5E2-4x%2Bc=0 and solve for x
x%5E2-4x%2B3=0
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-%28-4%29+%2B-+sqrt%28+%28-4%29%5E2-4%2A1%2A3+%29%29%2F%282%2A1%29+

x+=+%284+%2B-+sqrt%28+16-12+%29%29%2F2+
x+=+%284+%2B-+sqrt%28+4+%29%29%2F2+
x+=+%284+%2B-+2%29%2F2+


solutions:
x+=+%284+%2B2%29%2F2+
x+=+6%2F2+
x+=+3+
or
x+=+%284+-2%29%2F2+
x+=+2%2F2+
x+=+1+

+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E2-4x%2B3%29+

D=16+-+4c...set D equal to 0 and find c
0=16+-+4c
4c=16
c=4
do same for c=4 to prove that the equation has 1 real solution
D=16+-+4%2A4
D=16+-+16
D=0
+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E2-4x%2B4%29+

D=16+-+4c...set D greater then 0 and find c
0%3E16+-+4c
4c%3E16
c%3E4

do same for c%3E4 to prove that the equation has 2 conjugate imaginary solutions
take c=5
D=16+-+4%2A5
D=16+-+20
D=-4 => D%3C0
+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E2-4x%2B5%29+