Question 997637


Given the equation {{{4x^2+8x+k=0}}} , for what values of {{{k}}} will the equation have a {{{repeated}}} root (one double root), {{{2}}} real roots, and {{{3}}} values of k that will give the equation rational roots.

First of all, every polynomial has a discriminant, not just quadratics.  The discriminant is the first place you look to classify the types of roots a polynomial has.

For example,  a quadratic equation ax^2 + bx + c = 0, has a discriminant {{{b^2 - 4ac}}}

If {{{b^2 - 4ac > 0}}}, there are {{{two}}} real roots. 
If {{{b^2 – 4ac<0}}} there are two imaginary roots .
If {{{b^2 – 4ac=0}}}, there is {{{one}}} repeated root. 


so, in your case we have

{{{4x^2+8x+k=0}}} where {{{a=4}}}, {{{b=8}}}, and {{{c=k}}}

If {{{b^2 - 4ac > 0}}}, there are {{{two}}} real roots. 

so,

{{{b^2 - 4ac > 0}}}

{{{8^2 - 4*4k > 0}}}

{{{64- 16k > 0}}}

{{{64> 16k }}}

{{{64/16> k }}}

{{{4> k }}} or

{{{highlight(k<4)}}}

take first number less then {{{4}}}:{{{k=3}}} and check the roots

{{{4x^2+8x+3=0}}} ...when you check it, you will find out that roots are: {{{x = -3/2}}} and {{{x = -1/2}}}


{{{ graph( 600, 600, -10, 10, -10, 10, 4x^2+8x+3) }}}



If {{{b^2 – 4ac<0}}} there are two imaginary roots .

{{{b^2 – 4ac<0}}}

{{{8^2 - 4*4k < 0}}}

{{{64- 16k < 0}}}

{{{64<16k }}}

{{{64/16< k }}}

{{{highlight(k>4)}}}

take {{{k=5}}} and check


{{{ graph( 600, 600, -10, 10, -10, 10, 4x^2+8x+5) }}}


If {{{b^2 – 4ac=0}}}, there is {{{one}}} repeated root.


{{{b^2 – 4ac=0}}}

{{{8^2 - 4*4k =0}}}

{{{64- 16k = 0}}}

{{{64=16k }}}

{{{64/16= k }}}

{{{highlight(k=4)}}}

take {{{k=4}}} and check


{{{ graph( 600, 600, -10, 10, -10, 10, 4x^2+8x+4) }}}