Question 997637
4x^2+8x+k=0
4x^2+8x=-k
Factor out 4.
4(x^2+2x)= -k  but k can be any constant, so it doesn't have to be divided by 4.
If you complete the square, the roots will be the same, -1.
4(x^2+2x+1)= -(k)+4, adding 4 to both sides
4(x+1)^2= -k +4
The right side has to equal zero for completing the square to give two roots of -1
-k=-4
k=4
4x^2+8x+4=0
{{{graph(300,200,-10,10,-10,10,4x^2+8x+4)}}}
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Two real roots can be many things, so long as the discriminant b^2-4ac is positive.
b^2>4ak, since k is acting as c
64>16k
k<4 will work
{{{graph(300,200,-10,10,-10,10,4x^2+8x+2,4x^2+8x-2)}}}
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Rational roots work if the discriminant gives a perfect square, for all parts of the quadratic formula will be a fraction with nothing but integers.
b^2-16k has to be a perfect square
64-16k

k=0 works and roots are 4 and 0
k=3 :(1/8){-8 +/- sqrt(16){=(1/8)(-4) and (1/8)(-12)
k= -36/16 (1/8) (-8 +/- sqrt (100) and (1/8)(-8+10), (1/8) (-8-10) or (10/8), (-18/8)