how to prove the roots of a quadratic equation is always negative?

Question

Given k is a real constant such that 0 < k < 1
kx^2 + 2x + (1-k) = 0
show that the roots of the equation are always real and always negative.

Answer 1

the quadratic formula tells you that your roots will be

(-b+/- (b^2-4ac)^1/2)/2a

Ok. the denominator is always positive.

-b is -2. That's a negative number.

(b^2-4ac)=(2^2 - 4*k*(1-k))=4-4(k^2-k)

Since k is less than one, k^2-k can never be bigger in magnitude than -1/4. (This can be proven in many ways. Calculus is easiest but probably beyond this class. Try using 0, 1/2, 1 in the expression and see what you get.)

Thus (b^2-4ac) at it's biggest will be 3. Thus, the square root of 3 even when added to -b (which we said was -2) will be a negative number.

If the numerator is negative, and the denominator is positive, the fraction will end up negative.