Quadratic equation within quadratic

Question

Edited version:-

If both roots of equation $$4x^2 -20px+(25p^2 +15p- 66)=0 $$ are less than $2$, find the value of $p$. How to solve such type of equations?

Shailesh · Answer 1 · 2015-09-21T06:25:23.000

3

Hint :

Rewrite as $(2x - 5p)^2 = 66 - 15p$, so that $2x - 5p = \pm \sqrt{66-15p}$.

Now you want $66 - 15p \gt 0$ and you want $x_1$ and $x_2$ to be less than 2. I'm sure you can take it from here.

Another way would be to use the condition for both roots of a quadratic $ax^2 + bx + c = 0$ to be less than 2 are (1) Discriminant positive, (2) $ 2 \gt \frac{-b}{2a}$ and (3) $af(2) \gt 0$ where $f$ is the quadratic

edited Sep 21 '15 at 06:25

answered Sep 21 '15 at 06:17

Shailesh

3,789

You can use \pm to get the symbol $\pm$ in front of the square root, if you like. – String Sep 21 '15 at 06:21
1

@String Thanks. Have edited accordingly. – Shailesh Sep 21 '15 at 06:24
I think he wants $p < 2$ for both $x_1, x_2$ and not $x_1, x_2 < 2$ – Silver Sep 21 '15 at 06:25
There is certainly some ambiguity in the way the question is poorly stated... – Silver Sep 21 '15 at 06:27
Please will u solve further. – Muzamil Sep 21 '15 at 06:30
Also it's my first question here sorry for unclearness – Muzamil Sep 21 '15 at 06:32
Please send me a link by which I can understand the third alternative u stated above . – Muzamil Sep 21 '15 at 06:47
@Muzamil See [here] (http://www.askiitians.com/iit-jee-algebra/quadratic-equations-and-expressions/interval-in-which-the-roots-lie.aspx), though this does not show a graph, does take all cases. Please accept answer once you are satisfied. – Shailesh Sep 21 '15 at 06:51

Math-fun · Answer 2 · 2015-09-21T08:34:05.133

2

The roots are $$\frac{1}{2} \left(5 p\pm\sqrt{3} \sqrt{22-5 p}\right)$$ where clearly the biggest root is $\frac{1}{2} \left(5 p+\sqrt{3} \sqrt{22-5 p}\right)$. Now if this root is smaller than $2$ the other root will also be smaller than $2$. \begin{align} \frac{1}{2} \left(5 p+\sqrt{3} \sqrt{22-5 p}\right)&<2\\ 5 p+\sqrt{3} \sqrt{22-5 p}&<4\\ \sqrt{3} \sqrt{22-5 p}&<4-5p\\ 3 (22-5 p)&<(4-5p)^2\\ 66-15 p&<16 - 40 p + 25 p^2\\ 0&<-50 - 25 p + 25 p^2\\ 0&<-2 - p + p^2\\ 0&<(p+1)(p-2)\\ \end{align} If $p<-1$ or $p>2$ will the inequality hold ... (why?) Also note that $\sqrt{3} \sqrt{22-5 p}<4-5p$ requires $0<4-5p$ i.e. $p<\frac45$. Hence only $p<-1$ is acceptable. On the other hand we require $p<\frac{22}{5}$ hence $$p<-1.$$

edited Sep 21 '15 at 08:34

answered Sep 21 '15 at 07:40

Math-fun

9,507

Not true. Take p = 3. The equation is $4x^2 - 60x + (225+45-66) = 0$ which simplifies to $2x^2 - 30x + 102 = 0$ , where one root is almost near 10. – Shailesh Sep 21 '15 at 07:53
@Shailesh ($p<-1$ OR $p>2$) AND ($p<\frac{22}5$) is equivalent to ($p<-1$) OR ($2<p<\frac{22}5$). – Math-fun Sep 21 '15 at 08:09
I understand that, but when I take p = 3, I get one root which is greater than more than 2 which is not what the problem wants. So something has gone wrong somewhere – Shailesh Sep 21 '15 at 08:16
1

ups thanks for the comments. I just fixed the problem :-) – Math-fun Sep 21 '15 at 08:34
I'm not verifying it, but now I like your method better than the way I have written my answer. – Shailesh Sep 21 '15 at 08:39
Thanks a lot all but can we square both sides of inequalty? – Muzamil Sep 21 '15 at 16:31
@Muzamil as long as you know the sign of both sides yeah, for example $-6<1$ but $(-6)^2>1$ ... – Math-fun Sep 22 '15 at 12:51

Quadratic equation within quadratic

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