Q.1 Find the values of K so that the quadratic equationx^2+2(K+1)x+K+5=0 has at least one positive root.?

1 Answer
Aug 25, 2017

K=-5

Explanation:

A quadratic equation basic formula is
ax^2+bx+c
so matching up with x^2+2(K+1)x+k+5=0
gives
x^2 terms: x^2 :.a=1
x terms: 2(K+1) or 2K+2 :.b=2K+2
constants: K+5 :. c=K+5

For a quadratic to have at least one positive root, the discriminate (Delta) must be equal to (1 real root) or greater than (two real roots) 0.

From the quadratic formula, the discriminate is b^2-4ac. We need b^2≥4ac

So in this case, (2K+2)^2≥(4*1*(K+5))
:.4K^2+4K+4≥4K+20
K^2+K+1≥K+5
K^2+1≥+5
K≥4

So if K=4 then
x^2+2(4+1)x+(5+4)=0
x^2+10x+9=0
(x+9)(x+1) :. x=-9 and x=-1 These are both negative so K!=4

It seems like we need the c part to cancel. So if K=-5 then
the equation would be
x^2+2(-5+1)x+(5-5)=0
x^2-8x+0=0
x(x-8)
:. x=0 and x=8

8 is +ve so K=-5 must be the solution.