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max2010maxim [7]
3 years ago
9

HElp me with this one but not with a big processs justa little one

Mathematics
1 answer:
xeze [42]3 years ago
4 0
So 2 real solutions
just write 2 factors and expand
(x-3)(x-4)=0
x^2-7x+12=0
discriminant is when you have b^2-4ac
so subsitute
ax^2+bx+c=0
a=1
b=-7
c=12
-7^2-4(1)(12)
49-48=1
discriminant=1


one real solution would be
(x-1)(x-1)=0
x^2-2x+1=0
discriminant=-2^2-4(1)(1)=4-4=0
discriminant=0

no real solutions has the discriminant less than zero
let's say that b=1 and c=20 and a=1
basically, we need a set of numbers where b^2 is less than 4acso

1^2-4(1)(20)=1-80=-79
then we subsitute
x^2+x+20=0
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c + 3d = 14.75

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