Question #1cabf

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Question #1cabf

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Answer 1 · 2017-10-08T12:58:55

a) $k^{2} - 16 k + 60$ $k^2-16k+60$
b) $k = 6, 10$ $k=6, 10$
c) as in explanation

Explanation:

a) in $a x^{2} + b x + c = 0 (b^{2} - 4 a c) \Rightarrow {(- k - 8)}^{2} - 4 \cdot 1 \cdot (8 k + 1) = k^{2} - 16 k + 60$ $ax^2+bx+c=0 (b^2-4ac)=>(-k-8)^2-4*1*(8k+1)=k^2-16k+60$
$b) b^{2} - 4 a c = 0$ $b) b^2-4ac=0$ =>two equal roots=> $k^{2} - 16 k + 60 = 0 \Rightarrow (k - 6) (k - 10) = 0 \Rightarrow$ $k^2-16k+60=0=>(k-6)(k-10)=0=>$ k=6 or k=10
c) when $b^{2} - 4 a c < 0, f (x)$ $b^2-4ac<0, f(x)$ is always more than 0
so if $k = 8, b^{2} - 4 a c = - 4 < 0$ $k=8, b^2-4ac=-4<0$ , and $f (x) > 0$ $f(x)>0$ always (when $a > 0)$ $a>0)$

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Question #1cabf

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