How many real roots does the function f(x)=x^2+8x-2
2 answers:
Using the discriminant
for
f(x)=ax²+bx+c
the discriminant is b²-4ac
when it is
1. less than 0, there are no real roots
2. equal to 0, then there is 1 real root
3. more than 0, 2 real roots
given
f(x)=1x²+8x-2
a=1
b=8
c=-2
discriminat
8²-4(1)(-2)
64+8
this is greater than 0
2 real roots
First find the determinant, Δ = b^2 - 4ac
<span>a = 1, b = 8 , c = -2 </span>
<span>Δ = 8^2 - 4*1*-2 </span>
<span>Δ = 64 + 8 </span>
<span>Δ = 72 > 0 </span>
<span>therefore there are two real roots!</span>
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