Having Difficulty Answering These Questions- Can Someone Please Help Me? (1 Viewer)

181jsmith · Mar 14, 2012

See questions below, thanks

1. If a > o, b> 0 and c>0, show that $\frac{a+b+c}{3}\geq \sqrt[3]{abc}$

2. Show that, if $x^{3} + px + r = 0$ has a root of multiplicity two, then $27r^{2} + 4p^{3} = 0$

3. Show that $\frac{a^{2} + b^{2} + c^{2}}{3}\left\geq \left ( \left \frac{a + b + c}{3} \right )^{2}$

4. Find the roots of $\frac{3}2{}x^{4} - \frac{11}{2}x^{3} + 5x - 2=0$

5. In the diagram, the bisector of the $\angle RPQ$ meets RQ in S and the circum-circle of the triangle PQR in T.

Prove that $PS^{2}= PQ \times PR-RS\times SQ$

6.

AB is the common chord of two circles C1 and C2. AC and AD are chords of the respective circles with
$\angle CAB= \angle DAB$ and CD meets the circles at P and Q respectively. R is the foot of the perpendicular from P to BQ. Prove that $\angle PBQ= 2\angle RPQ$

q 5 and 6 diagrams : View attachment 24582 View attachment 24581

nightweaver066 · Mar 14, 2012

$2. $Let $ P(x) = x^3 + px + r$

$P'(x) = 3x^2 + p$

$By multiple root theorem, if it has a root of multiplicity two, $ P(\alpha) = P'(\alpha) = 0 $ where $\alpha $is the double root$

$P'(\alpha) = 3\alpha^2 + p = 0$

$\alpha = \pm\sqrt{\frac{-p}{3}}$

$P(\alpha) = \alpha^3 + p\alpha + r = 0$

$-\frac{p}{3}(\pm\sqrt{\frac{-p}{3}}) + p(\pm\sqrt{\frac{-p}{3}}) + r = 0$

$(p - \frac{p}{3})(\pm\sqrt{\frac{-p}{3}}) = -r$

$Square both sides and simplify$

$4. \frac{3}{2}x^4 - \frac{11}{2}x^3+ 5x - 2 = 0$

$3x^4 - 11x^3 + 10x - 4 = 0$

$$Test x = \frac{2}{3} $, 3(\frac{2}{3})^4 - 11(\frac{2}{3})^3 + 10(\frac{2}{3}) - 4$

$= 0$

$\therefore $ (3x - 2) is a factor by factor theorem$$

$(3x - 2)(x^3 - 3x + 2) = 0$

$(3x - 2)(x - 1)(x^2 + x - 2) = 0$

$(3x -2)(x - 1)^2(x + 2) = 0$

$\therefore x = \frac{2}{3}, 1, -2$

Carrotsticks · Mar 14, 2012

Geez you don't ask for much do you? =p

Don't think I'll be able to provide ALL solutions for you, but I'll provide a few quick points to get you started.

1. Look into the 'AM-GM Inequality'. Your expression is the equivalent for n=3.

2. Find the y-coordinates of the stationary points and let it be equal to zero (for the stationary points to be on opposite sides of the x axis). Then once you simplify that expression, you should get what you need. Except realise that equality occurs when it has a root of multiplicity. That expression there is also the 'discriminant' for cubics.

3. If I recall correctly, the Terry Lee book has the same (or similar) question. Might want to check it out.

4. By observing it the function, I can see that x=-1 is a solution. So I factor out (x+1) as a factor and this leaves me a cubic. It is clear that this cubic has a root x=2/3 so I factor out (3x-2), leaving a quadratic. Solve this quadratic to acquire the other 2 solutions.

181jsmith · Mar 15, 2012

thanks for your answer nightweaver- really appreciate it. do u mind telling me how u know how to test x=2/3? and how you get $x^{3}-3x+2$ as a factor? - i'm just a bit lost.

181jsmith · Mar 15, 2012

thanks for ur answer carrotsticks- really appreciate it as well. Do you mind explaining more about the AM-GM Inequality you were
mention ing Q1?i'm just a bit confused. also, would u have any idea of which book some of the other questions are from? becos u said Q3 seemed to be from Terry Lee, thanks

Carrotsticks · Mar 15, 2012

Q1 is in T.Lee too I think under "Harder 3U Inequalities"

Q2 is in T.Lee as well under "Polynomials"

The AM-GM inequality follows as such:

$\frac{x_1+x_2+x_3+...+x_n}{n} \geq \sqrt[n]{x_1 x_2 x_3 ... x_n}$

Question 1 is the same thing, but only going up to x_3.

And instead of x_1, x_2 etc etc, they used a, b ...

181jsmith · Mar 17, 2012

Carrotsticks said:
Q1 is in T.Lee too I think under "Harder 3U Inequalities"

Q2 is in T.Lee as well under "Polynomials"

The AM-GM inequality follows as such:

$\frac{x_1+x_2+x_3+...+x_n}{n} \geq \sqrt[n]{x_1 x_2 x_3 ... x_n}$

Question 1 is the same thing, but only going up to x_3.

And instead of x_1, x_2 etc etc, they used a, b ...

ah great, that's a big help,btw for q4- i get a quadratic of $x^{2}-4x-2$ using the steps u provided me with, where would i go from there to get the roots? thanks heaps

dulip · Mar 17, 2012

quadratic formula or completing the square to find the other two roots

Having Difficulty Answering These Questions- Can Someone Please Help Me? (1 Viewer)

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