Question

x^2+kx+k+1=0

a) for what values of k will the equation have two real roots?
b) fin the largest whole number value of k for which the equation will have two complex solution
c) solve the equation for the value of k found in part (b)

Answer 1

easy

Answer 2

discriminant should be greater than o0 to have 2 roots ..equate this and u get value of k

Answer 3

i see k can can have four values here

Answer 4

see here

Answer 5

http://www.wolframalpha.com/input/?i=x^2%2Bkx%2Bk%2B1%3D0

Answer 6

all efforts and no medal?

Answer 7

\[\Delta = k^2 -4(1)(k+1)\]

Answer 8

Now we need this discriminant to be POSITIVE for two real roots.

It is a quadratic expression so check the discriminant of the discriminant that we just calculated.

\[\Delta2= (-4)^2 -4(1)(-4) = 16+16 >0 \]

Answer 9

So the original eqn has two real roots for all k.

That can't seem correct given it has extra parts.

Answer 10

In this case a=1,b=k,c=k+1, \[\Delta=k ^{2}-4\left( k +1 \right)=k^{2}-4k -4\]\[k=\left( 4\pm \sqrt{16-4\times -4} \right)\div2\]\[k=\left( 4\pm \sqrt{32} \right)\div2=\left( 4\pm4\sqrt{2} \right)\div2=2\pm2\]