multiple zeroes question (1 Viewer)

saltshaker · Sep 14, 2021

Screen Shot 2021-09-14 at 7.45.22 pm.png

When I sub in x = -2 on the original eqn m gets deleted?

CM_Tutor · Sep 14, 2021

saltshaker said:
When I sub in x = -2 on the original eqn m gets deleted?

Yes because $x=-2$ is a root of the equation.

Remember that the roots are $x = -2,\ \alpha,\ \text{and}\ \alpha$ , so you know that the sum of the roots is $2\alpha - 2 = ...$

saltshaker · Sep 14, 2021

this topic will singlehandedly drop my avg from 100% to 50%

ty

saltshaker · Sep 14, 2021

Screen Shot 2021-09-14 at 8.44.36 pm.png

Just (b)
Ans (k =0 T(origin), k = 4/27 T(2/3, 8/27)

The question doesn't make any sense anyways but idk if I'm missing something

Lith_30 · Sep 14, 2021

part b:
You have to solve the two equations simultaneously so that any points of intersection will be shown as the zeroes of the resultant graph

$\\y=x^3\\y=x^2-k\\\text{sub two equations together}\\x^3=x^2-k\\x^3-x^2+k=0$

I think the hardest part of the question is finding out how many points of intersection there will be. Since the the parabola is meant to touch the cubic there has to be a double root, but looking at the equation we just got there is one more root as cubic equations have three roots.

Let these roots be $x=\alpha,\alpha,\beta$

sum of roots
$2\alpha+\beta=1\longrightarrow \textcircled{1}$

sum of product of pairs
$2\alpha\beta+\alpha^2=0\longrightarrow \textcircled{2}$

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

From $\textcircled{1}$ we can rearrange to make $\beta$ the subject.
$\beta=1-2\alpha\longrightarrow \textcircled{4}\$

sub $\textcircled{4}$ into $\textcircled{2}$

$\\2\alpha(1-2\alpha)+\alpha^2=0\\2\alpha-3\alpha^2=0\\-\alpha(3\alpha-2)=0\\\alpha=0,\ \alpha=\frac{2}{3}$ now we sub these into $\textcircled{4}$ to get the values of $\beta$ .

at $\alpha=0$
$\\\beta=1-2(0)\\\beta=1$

at $\alpha=\frac{2}{3}$
$\\\beta=1-2 \left (\frac{2}{3} \right )\\\beta=-\frac{1}{3}$

Now we sub the two different sets of values for $\alpha$ and $\beta$ into $\textcircled{3}$ to find k

$\\(0^2)(1)=-k\\k=0\\\\\\ \left (\frac{2}{3} \right )^2 \left (-\frac{1}{3} \right )=-k\\k=\frac{4}{27}$

Now that we have the two values of k, we now need to find the two different points where the two original graphs touch, which can be done by subbing in the values of $\alpha$ into $y=x^3$ , and that's where we get the two different coordinates.

CM_Tutor · Sep 14, 2021

If $y=x^2-k$ and $y=x^3$ "touch" then the equation $x^2-k=x^3$ must have a double root.

Let's put that double root at $x=\alpha$ and let $x=\beta$ be the other root for this equation, which can be expressed as $x^3-x^2+k=0$ . From root theory:

$\begin{align*} \text{Sum of Roots}: \qquad 2\alpha + \beta &= -\cfrac{-1}{1} = 1 \qquad \text{. . . . . . . . . . (1)} \\ \text{Sum of Roots in Pairs}: \qquad \alpha^2 + 2\alpha\beta &= \cfrac{0}{1} = 0 \qquad \text{. . . . . . . . . . (2)} \\ \text{Product of Roots}: \qquad \alpha^2\beta &= -\cfrac{k}{1} = -k \qquad \text{. . . . . . . . . . (3)} \end{align*}$

From (2), we see that $\alpha(\alpha + 2\beta) = 0$ , giving two solutions:

$\textbf{Solution 1:} \quad \alpha = 0 \ \implies \ \beta = 1 \text{ from (1), and, from (3), } k = 0$
$\text{The curves touch at $x = \alpha = 0$, i.e at $T(0,\ 0)$, and cross when $x=\beta = 1$, i.e. at $(1,\ 1)$.}$

$\textbf{Solution 2:} \quad \alpha = -2\beta \ \implies \ \beta = -\cfrac{1}{3} \text{ from (1), and, from (3), } k = \cfrac{4}{27}$
$\text{The curves touch at $x = \alpha = \cfrac{2}{3}$, i.e at $T\left(\cfrac{2}{3},\ \cfrac{8}{27}\right)$, and cross when $x=\beta = -\cfrac{1}{3}$, i.e. at $\left(-\cfrac{1}{3},\ -\cfrac{1}{27}\right)$.}$

=)(= · Sep 16, 2021

CM_Tutor said:
Yes because $x=-2$ is a root of the equation.

Remember that the roots are $x = -2,\ \alpha,\ \text{and}\ \alpha$ , so you know that the sum of the roots is $2\alpha - 2 = ...$

wait how did you find the equation of l

Lith_30 · Sep 16, 2021

=)(= said:
wait how did you find the equation of l

You use the point gradient formula $y-y_1=m(x-x_1)$

=)(= · Sep 16, 2021

ohh okay, how would you find m

Lith_30 · Sep 16, 2021

=)(= said:
ohh okay, how would you find m

You get the value of m by subbing the equation of $l$ into $y=x^3-x+3$ , then using the products and sum of roots to find the value of m.

=)(= · Sep 16, 2021

Lith_30 said:
You get the value of m by subbing the equation of $l$ into $y=x^3-x+3$ , then using the products and sum of roots to find the value of m.

Sorry would you be able to type the working out please?

CM_Tutor · Sep 16, 2021

We know that, for any cubic equation $Ax^3+Bx^2+Cx+D=0$ with roots $\alpha$ , $\beta$ , and $\gamma$ , the roots are related by three equations:

$\text{Sum of Roots:} \qquad \alpha + \beta + \gamma = -\cfrac{B}{A}$

$\text{Sum of Roots in Pairs:} \qquad \alpha\beta + \beta\gamma + \gamma\alpha = \cfrac{C}{A}$

$\text{Product of Roots:} \qquad \alpha\beta\gamma = -\cfrac{D}{A}$

Applying this to the equation $x^3-(m+1)x+(6-2m)=0$ with roots of $-2,\ a,\ \text{and}\ a$ , we get:

$\begin{align*} \text{Sum of Roots:} \qquad -2 + a + a &= -\cfrac{0}{1} \\ 2a - 2 &= 0 \\ a &= 1 \end{align*}$

$\begin{align*} \text{Sum of Roots in Pairs:} \qquad -2a + a^2 + a(-2) &= \cfrac{-(m+1)}{1} \\ a^2 - 4a &= -(m + 1) \\ 1 - 4 &= -(m + 1) \qquad \text{as $a = 1$.} \\ m &= 2 \end{align*}$

$\begin{align*} \text{Product of Roots:} \qquad -2a^2 &= -\cfrac{6-2m}{1} \\ a^2 &= 3 - m \qquad \text{which is consistent with $a=1$ and $m=2$} \end{align*}$

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multiple zeroes question (1 Viewer)

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