Given that the equation, $(k-1)x^2-2(k-1)x-(3k+1)=0$ has real roots, show that $k^2-k≥0$

Question

I can get to $k^2-k≥0$ but only when I make $b^2$ negative. The problem is why would I make $b^2$ negative other than the fact that $b$ is negative in the original equation? The problem with this is that $c$ is also negative and so I would also have to make c negative which gives me $8x^2-8≥0$

Help me please, regards

More information: This is Question 9(b)(ii) from Past Exam Paper AQA Mathematics A Level Unit Further Pure 1 June 2013

Answer 1

You have to check if the discrimiant is positive. The discriminant of $ax^2 + bx +c$ is $b^2 -4ac$. In your case $a= (k-1)$, $b = -2(k-1)$ and $c = -(3k+1)$.

Thus you need to check if the following is positive: $$ (-2(k-1))^2 - 4(k-1)(-(3k+1)) = (2(k-1))^2 + 4(k-1)(3k+1).$$ Simplifying this will yield something directly equivalent to your condition.

Pay attention that the formula is given for $ax^2 + bx +c$, if you have minuses in the expression they need to be taken into account when determining the $a,b,c$. In case you find this difficult, rewrite the same expression introducing the pluses to make this more clear.

Answer 2

A quadratic equation with real roots has discriminant $ \geq 0$. Here we have $$ 0 \leq D = 4(k-1)^2 + 4 (k-1)(3k+1) = 4(k-1)(4k) = 16k(k-1), $$ implying $k^2 - k = k(k-1) \geq 0$.

Answer 3

Note that $k=1$ would turn your equation into $-4=0$ for all $x$ so we may suppose that $k\neq 1$. In this case, the given equation is equivalent to: $$ 0=x^2-2x-\frac{3k+1}{k-1}=x^2-2x-\frac{-k+1+4k}{k-1}=(x-1)^2-\frac{4k}{k-1}. $$ This is solvable for real $x$ iff $4k/(k-1)\geq 0$. Multiplying this last inequality by $\frac{(k-1)^2}{4}>0$ gives the desired result.

Answer 4

Calculate its discriminant and set it to zero.

$ 4(k-1)^2 + 4 (k-1)(3k +1) = 4\,(k-1)*4k = 16\,k\,(k-1) $

Graph of $ y(k) = k(k-1) $ is a parabola with roots k = 0,1. The portion $ 0 < k < 1 $ lies above x-axis and that part only gives real roots.