Pls help (1 Viewer)

ExtremelyBoredUser · Oct 1, 2021

Could you give the whole question? This seems like it requires prior context to do since in theory a regular parabola within a certain domain for k would suffice?

=)(= · Oct 1, 2021

ExtremelyBoredUser said:
Could you give the whole question? This seems like it requires prior context to do since in theory a regular parabola within a certain domain for k would suffice?

CM_Tutor · Oct 1, 2021

The only way for a parabola to touch but not cross a circle is if the two have common tangents at each of the points of intersection, and thus a double root where the curves meet. So, the question is asking for $k$ such that $x^2 + \left(x^2 - k\right)^2 = 1$ , a quartic, has two double roots.

Lith_30 · Oct 1, 2021

First we have to solve the two equations simultaneously so that the zeros of that equation are the points of intersection.

$\\y=x^2-k\longrightarrow\textcircled{1}\\x^2+y^2=1\longrightarrow\textcircled{2}$

Sub $\textcircled{1}$ into $\textcircled{2}$

$\\x^2+(x^2-k)^2=1\\x^2+x^4-2x^2k+k^2=1\\x^4+(1-2k)x^2+(k^2-1)=0$

Since the two graphs touch each other at two points, there will be two double roots which we'll label as $\alpha,\ \alpha,\ \beta,\ \beta$

Sum of roots:
$2\alpha+2\beta=0\longrightarrow\textcircled{1}$

Sum of product of pairs:
$\alpha^2+4\alpha\beta+\beta^2=1-2k\longrightarrow\textcircled{2}$

Product of roots:
$\alpha^2\beta^2=k^2-1\longrightarrow\textcircled{3}$

From $\textcircled{1}$ we get $\beta=-\alpha$ , sub that into $\textcircled{2}$

$\\\alpha^2-4\alpha^2+\alpha^2=1-2k\\\\\frac{-2\alpha^2-1}{-2}=k\\\\k=\frac{2\alpha^2+1}{2}\longrightarrow\textcircled{4}$

Sub $\textcircled{4}$ and $\beta=-\alpha$ into $\textcircled{3}$

$\\\alpha^2(-\alpha)^2=\left (\frac{2\alpha^2+1}{2} \right )^2-1\\\\\alpha^4=\left (\frac{4\alpha^4+4\alpha^2+1}{4} \right)-1\\\\\alpha^4=\alpha^4+\alpha^2+\frac{1}{4}-1\\\\\alpha^2-\frac{3}{4}=0$
$\alpha=\frac{0\pm\sqrt{0-4(-\frac{3}{4})}}{2}\text{Using quadratic formula}$
$\\\alpha=\frac{\sqrt{3}}{2},\ \alpha=-\frac{\sqrt{3}}{2}\\\\\therefore\alpha=\frac{\sqrt{3}}{2},\ \beta=-\frac{\sqrt{3}}{2}\ \text{or}\ \alpha=-\frac{\sqrt{3}}{2},\ \beta=\frac{\sqrt{3}}{2}$

To find the value of k we can just sub alpha and beta into $\textcircled{3}$

$\\\left(\frac{\sqrt{3}}{2}\right)^2+4\left(\frac{\sqrt{3}}{2}\right)\left(-\frac{\sqrt{3}}{2}\right)+\left(-\frac{\sqrt{3}}{2}\right)^2=1-2k\\\\\frac{3}{4}-3+\frac{3}{4}=1-2k\\\\k=\frac{-\frac{3}{2}-1}{-2}\\\\k=\frac{5}{4}$

Now to find the points at which the graph touch each other we just sub the values of alpha and beta into the original equation.

$\\\text{at}\ \alpha=\frac{\sqrt{3}}{2}\\\\y=\left(\frac{\sqrt{3}}{2}\right)^2-\frac{5}{4}\\\\y=-\frac{1}{2}$

$\\\text{at}\ \beta=\frac{-\sqrt{3}}{2}\\\\y=\left(-\frac{\sqrt{3}}{2}\right)^2-\frac{5}{4}\\\\y=-\frac{1}{2}$

Therefore points of contact are $\left(\frac{-\sqrt{3}}{2},\ -\frac{1}{2}\right)\ \text{and}\ \left(\frac{\sqrt{3}}{2},\ -\frac{1}{2}\right)$ and $k=\frac{5}{4}$

=)(= · Oct 1, 2021

Thank you so much

I have another question

jimmysmith560 · Oct 1, 2021

=)(= said:
Thank you so much

I have another question
View attachment 32381

An identical question (with a couple of different values) was previously asked. It was as follows:

Two roots of $x^3+mx^2+15x-7=0$ are equal and rational. Find m.

The working out for this question is included below:

You should use this as guidance to solve your question (i.e. using the appropriate values from your own question).

Important edit:

As @CM_Tutor kindly pointed out, the working out attached should actually say to look at the roots of 14, which are

, and not the roots of 34. Further, the roots of 34 are incorrectly stated, as they should be

. Apologies for the typos!

I hope this helps!

CM_Tutor · Oct 1, 2021

Lith_30 said:
First we have to solve the two equations simultaneously so that the zeros of that equation are the points of intersection.

$\\y=x^2-k\longrightarrow\textcircled{1}\\x^2+y^2=1\longrightarrow\textcircled{2}$

Sub $\textcircled{1}$ into $\textcircled{2}$

$\\x^2+(x^2-k)^2=1\\x^2+x^4-2x^2k+k^2=1\\x^4+(1-2k)x^2+(k^2-1)=0$

I suspect it is more efficient to recognise this as a quadratic in $x^2$ and solve it directly as:

$\begin{align*} x^2 &= \cfrac{-(1-2k) \pm \sqrt{(1-2k)^2 - 4(1)(k^2 - 1)}}{2(1)} \\ &= \cfrac{2k - 1 \pm \sqrt{1-4k + 4k^2 - 4k^2 + 4}}{2} \\ &= \cfrac{2k - 1 \pm \sqrt{5 - 4k}}{2} \end{align*}$

Now, if there is a solution to this equation with $x^2 < 0$ then we have complex roots and hence there aren't two double roots. If there are two roots with $x^2 > 0$ then we have four roots and thus no double roots. So, we need there to be exactly one value of x^2 and it can't be zero (draw a sketch to see why). For there only to be a single value of $x^2 > 0$ , we require:

$5 - 4k = 0 \qquad \implies \qquad k = \cfrac{5}{4}$

$k = \cfrac{5}{4} \qquad \implies \qquad x^2 = \cfrac{2k - 1 \pm \sqrt{5 - 4k}}{2} = \cfrac{2\left(\cfrac{5}{4}\right) - 1 \pm \sqrt{0}}{2} = \cfrac{3}{4}$

$x^2 = \cfrac{3}{4} \qquad \implies \qquad y = x^2 - k = \cfrac{3}{4} - \cfrac{5}{4} = -\cfrac{1}{2}$

$\text{The points we seek occur when}\ k = \cfrac{5}{4}\ \text{and when}\ x = \pm \cfrac{\sqrt{3}}{2},\ \text{and so, are located at}\ \left(\cfrac{\sqrt{3}}{2},\ -\cfrac{1}{2}\right)\ \text{and}\ \left(\cfrac{\sqrt{3}}{2},\ -\cfrac{1}{2}\right).$

Perhaps more elegantly, examining $x^4+(1-2k)x^2+(k^2-1)=0$ , we can see that it must factorise as $\left(x^2 - A\right)\left(x^2 - B\right)$ . This further factorises, provided $A,\ B > 0$ , as $\left(x + \sqrt{A}\right)\left(x - \sqrt{A}\right)\left(x + \sqrt{B}\right)\left(x - \sqrt{B}\right)$ , and seen that it can only give two double roots if $A = B$ , and so note that $\Delta = 5 - 4k = 0$ .

CM_Tutor · Oct 1, 2021

jimmysmith560 said:
--- Snip --- working leading to $\alpha^3 - 15\alpha + 14 = 0$

View attachment 32383

The attachment should actually say to look at the roots of 14, which are $\pm 1,\ \pm 2,\ \pm 7,\ \text{and}\ \pm 14$ , and not the roots of 34. Further, the roots of 34 are incorrectly stated, as they should be $\pm 1,\ \pm 2,\ \pm 17,\ \text{and}\ \pm 34$ . Likely a couple of typos here.

jimmysmith560 · Oct 1, 2021

CM_Tutor said:
The attachment should actually say to look at the roots of 14, which are $\pm 1,\ \pm 2,\ \pm 7,\ \text{and}\ \pm 14$ , and not the roots of 34. Further, the roots of 34 are incorrectly stated, as they should be $\pm 1,\ \pm 2,\ \pm 17,\ \text{and}\ \pm 34$ . Likely a couple of typos here.

Thank you for pointing this out! I have adjusted my post to reflect this

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Pls help (1 Viewer)

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