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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

Thomas Winkworth
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4 Answers4

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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 $k^2 - k = k(k-1) \geq 0$.

user133281
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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.

Kim Jong Un
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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.

quid
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  • So -(2(k-1)) is not the same as (-2(k-1)) ? I think I am getting confused by the square function and how it actually applies to that negative – Thomas Winkworth Jan 31 '15 at 15:28
  • No $-(2(k-1))$ is the same as $(-2(k-1))$. What I meant to highlight is that the squaring makes the sign disapear. So for $b$ it would not even change anything if in error you took $b=2(k-1)$. But the error of taking $c= 3k-1$ has an effect. – quid Jan 31 '15 at 15:33
  • Having looked at your other comment it might be that you calculated the square of $-2(k-1)$ in a wrong way as $-4(k-1)^2$, yet it is $4(k-1)^2$. Note that $b^2 = (-b)^2$ yet $-b^2$ is something else. In now think this is what you meant in your first comment here. For example, $3^2 = (-3)^2 = 9$ but $-3^2 = -9$. So while $(-3)$ and $-3$ are the same thing. $(-3)^2$ and $-3^2$ are not. – quid Jan 31 '15 at 15:36
  • Something which is also confusing me, why is c=-(3k-1) rather than c=-(3k+1) – Thomas Winkworth Jan 31 '15 at 16:45
  • Sorry this was my error! I correct it. – quid Jan 31 '15 at 16:46
  • Can you show your working? How does the -4(k−1)(−(3k+1))=4(k−1)(3k+1) – Thomas Winkworth Jan 31 '15 at 16:52
  • Sure. It is basically that $-(-x)=x$, note $-4(k−1)(−(3k+1))= -4(k−1)(−1)(3k+1) = -(−1)4(k−1)(3k+1) =4(k−1)(3k+1)$ – quid Jan 31 '15 at 16:54
  • Shouldn't it be -4(k-1)((-1)(3k+1))? In which case you would have to multiply ((-1)(3k+1)) first? – Thomas Winkworth Jan 31 '15 at 16:57
  • Multiplication is associative so this does not make an actual difference. One thus also does not typically write all the parenthesis only those that are relvant. – quid Jan 31 '15 at 17:00
  • If -4(k−1)(−(3k+1))=4(k−1)(3k+1) then does this also mean -4(k−1)(−(3k+1))=-4(k−1)(3)(−(k+1)) considering -4(k-1)((-1)(3k+1))=-4(k-1)(-1)(3k+1) – Thomas Winkworth Jan 31 '15 at 18:38
  • No. You because $3k+1$ is not the same as $3(k+1)$ the latter is $3k + 3$ by the distributive law. – quid Jan 31 '15 at 18:45
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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.

Narasimham
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