For which value of k will the roots of $2x^2+kx+1=0$ be real?

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Answer 1 · 2016-12-02T20:56:36

The answer is $k in ] -oo,-sqrt8 ] uu [sqrt8, oo[$

In order for the roots of a quadratic equation, to de real, the dircriminant $Delta>=0$

If the equation is $ax^2+bx+c=0$

the discriminant is $Delta=b^2-4ac$

$2x^2+kx+1=0$

$Delta= k^2-4*2*1>=0$

$k^2-8>=0$

$(k-sqrt8)(k+sqrt8)>=0$

We do a sign chart

$color(white)(aaaa)$ $k$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-sqrt8$ $color(white)(aaaa)$ $sqrt8$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $k+sqrt8$ $color(white)(aaaa)$ $-$ $color(white)(aaaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $k-sqrt8$ $color(white)(aaaa)$ $-$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $Delta$ $color(white)(aaaaaaaa)$ $+$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $+$

So, for $Delta>=0$

$k in ] -oo,-sqrt8 ] uu [sqrt8, oo[$

For which value of k will the roots of 2x^2+kx+1=02x^2+kx+1=0 be real?