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If one solution of the equation $3x^2 = 8x + 2k + 1$ is $7$ times the other. Find the solutions and the value of $K$.

Note: This isn't a homework question. I'm skipping ahead in my textbook. Thank you for the help.

Also please help me with this question no.2 also. Find the other zeroes of $2x^4 - 3x^3 -3x^2 + 6x - 2$ if $-\sqrt{2}$ and $\sqrt{2}$ are two zeroes of the given polynomial

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Looks like today is really a "-b/a" and "c/a" day for you! – Zeta.Investigator Sep 1 '12 at 11:01
It pays off when the teacher introduces a new concept and you have no problems with it while others struggle to grasp it – Aayush Agrawal Sep 1 '12 at 11:09
If $-\sqrt{2}$ and $\sqrt{2}$ are two zeroes, then the polynomial is divisible by $x^2-2$. – Julian Kuelshammer Sep 1 '12 at 11:28
That part i figured out, but i cant get very far from there. – Aayush Agrawal Sep 1 '12 at 11:35
up vote 6 down vote accepted

$\bigstar$ First question:
First we form the standard notation:$3x^2-8x-(2k+1)=0$
We know that the sum of the roots of a quadratic equation of the form $ax^2+bx+c$ is $x_1+x_2=-\frac{b}{a}$
So: $x_1+x_2 =\frac{8}{3}$
On the other hand: $x_1=7\times x_2$
Solving this two equations two unknowns yields:$x_2 = \frac13$ and $x_1=\frac73$
We also know that the product of the roots of a quadratic equation of the form $ax^2+bx+c$ is : $x_1 \times x_2=\frac{c}{a}$
so $$x_1 \times x_2 = \frac79 = -\frac{(2k+1)}{3} \implies k=-\frac 53 $$

$\bigstar$ Second question:
We can write a polynomial as: $P(x)=(x-x_1)(x-x_2)...(x-x_n)$ where $x_1$ to $x_n $ are the zeros of $P(x)$
So given two roots we have: $$\begin{align} P(x)&=2x^4 - 3x^3 -3x^2 + 6x - 2\\ &=(x-\sqrt2)(x+\sqrt2)(x-x_1)(x-x_2)\\ &=(x^2-2)(x-x_1)(x-x_2) \end{align}$$ Dividing P by $x^2-2$ yields: $$\begin{align} P(x)&=(x^2-2)(2x^2-3x+1)\\ &=(x^2-2)(x-1)(2x-1) \end{align}$$ So:
$$x_1=1 , x_2=0.5$$

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Thanks, that explained alot! Also can you take a look at the second one too? Thanks! – Aayush Agrawal Sep 1 '12 at 11:10
@Gigili: I appended your edit.tnx – Zeta.Investigator Sep 1 '12 at 11:42

Rearranging the equation, $3x^2-8x-(2k+1)=0$

If $a,7a$ are the solutions, $a+7a=\frac{8}{3}\implies a=\frac{1}{3}$

So, $a\cdot 7a=-\frac{2K+1}{3}\implies \frac{7}{9}=-\frac{K+1}{3}\implies K=-\frac{5}{3}$

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Hint: Study this link. Vieta's formulas may prove useful later in your mathematical life.

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Thanks, i will take a look at that. Indian education system is hugely strict though, if you use a concept that isnt in your book you get a zero D: – Aayush Agrawal Sep 1 '12 at 11:15
Ok I see, for a second degree polynomial over the real numbers (say) the formulas are not that hard to derive. Assume $r_1$ and $r_2$ are the roots of a monic polynomial $p(x)$ over $\mathbb{R}$ then $p(x)=(x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2$. (Why?) Now it is just a matter of comparing coefficients. – Henrik Sep 1 '12 at 11:30

Hint $\ $ Vieta $\Rightarrow$ roots of $\rm\:x^4 \color{#0A0}{-3/2}\, x^3 - 3/2\, x^2 + 3\,x \color{#C00}{- 1} = 0\:$ have sum $\,\color{#0A0}{3/2},\:$ and product $\,\color{#C00}{-1}.\:$ Therefore $\rm\: \sqrt{2} -\sqrt{2} + r + s = 3/2,\,$ and $\rm\, -\sqrt{2}\,\sqrt{2}\, r\,s = -1,\ $ so $\rm\ r+s =\color{#0A0}{ 3/2},\,$ and $\rm\, r\,s = \color{blue}{1/2}.\:$ Therefore, applying Vieta again, $\rm\:r,s\:$ are the roots of $\rm\:x^2 - \color{#0A0}{3/2}\, x + \color{blue}{1/2} = 0,\:$ which are $\rm\:1,\ 1/2.$

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