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Solve: The equation $7x^2-(R+13)x+R^2-R-2=0$ has one real root in range (0,1) and one real root in range (1,2). What's the range of R?

I've been given this problem for a math class (the topic was discriminants) and I'm not really sure how to approach it. I took the discriminant in R and x, but I don't know how to use the information about the range.

Silverleaf1
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  • Think about this. For this to happen, the function must be positive on $(-\infty,1]\cup[2,\infty)$, and be negative somewhere on both $(0,1)$ and $(1,2)$. Knowing this is a quadratic, what do you know about the roots? – Rushabh Mehta Sep 22 '19 at 13:08

2 Answers2

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Take $f(x)=7x^2-(R+13)x+R^2-R-2$ as a function who has two roots, one in $(0,1)$ and one in $(1,2)$
so there must be simultanously $$f(0)\times f(1)<0\\and\\f(1)\times f(2)<0$$ this mean $$f(0)\times f(1)=(7(0)^2-(R+13)(0)+R^2-R-2)(7(1)^2-(R+13)1+R^2-R-2)<0\\ and\\ f(1)\times f(2)=(7(1)^2-(R+13)1+R^2-R-2)(7(2)^2-(R+13)2+R^2-R-2)<0$$ $$(R^2-R-2)(7-(R+13)+R^2-R-2)<0\\ and\\ (7-(R+13)+R^2-R-2)(28-2(R+13)+R^2-R-2)<0$$

Khosrotash
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Hint: $$x_1=\frac{1}{14} \left(-3 \sqrt{-3 R^2+6 R+25}+R+13\right)$$ $$x_2=\frac{1}{14} \left(3 \sqrt{-3 R^2+6 R+25}+R+13\right)$$

Dr. Sonnhard Graubner
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