SOLUTION: Hi! I need to calculate the discriminant and state how many real solutions each equation has. 4x^2-6x+4=0 and 4x^2-4x+1=0 Thanks for youe help in advance!

Question 66523: Hi! I need to calculate the discriminant and state how many real solutions each equation has.
4x^2-6x+4=0
and
4x^2-4x+1=0
Thanks for youe help in advance!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
calculate the discriminant and state how many real solutions each equation has.
4x^2-6x+4=0
a=4, b=-6, c=4
Discriminant= b^2-4ac = 6^2-4(4)(4)=36-64<0
Since it is <0 there are two complex roots, no Real Number solutions.
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and
4x^2-4x+1=0
a=4, b=-4, c=1
Discriminant= b^2-4ac = 16-4*4*1=0
Since it is 0 there is one Real Number solution with multiplicity two.
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Cheers,
Stan H.

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