Equation

**MichaelLight** · February 26th 2011, 05:55 PM

For question 21(1), i tried many times but i couldn't show it, after substituting everything, the equation appears to be really complicated, i wonder can anyone help me with 21(i)? Thanks in advance...

**tonio** · February 26th 2011, 06:35 PM

Originally Posted by MichaelLight

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For question 21(1), i tried many times but i couldn't show it, after substituting everything, the equation appears to be really complicated, i wonder can anyone help me with 21(i)? Thanks in advance...

Use Viette's formulae: " the product of both roots of a quadratic eq. equals its free coefficient divided by its leading coef.,

and the sum of the roots equals minus its linear coef. divided by its leading one".

Thus, $\displaystyle{\frac{m^2}{ln}=\frac{\left(l\lambda( \lambda+1)^2+(\lambda+1)^2\right)^2}{ln}=\frac{l(\ lambda+1)^6}{n}}$

$\displaystyle{\frac{b^2}{ac}=\frac{\left(-a(\alpha+\beta)\right)^2}{a^2\alpha\beta}=\frac{(\ alpha+\beta)^2}{\alpha\beta}=\frac{(\lambda\beta+\ beta)^2}{\lambda\beta^2}=\frac{(\lambda+1)^2}{\lam bda}}$ .

Well, now just show that both expressions in the RHS above are equal (Viete again),

and for part (ii) is similar.

Tonio

**topsquark** · February 26th 2011, 06:42 PM

$\displaystyle ax^2 + bx + c = 0$

Thus
$\displaystyle a \left ( x^2 + \frac{b}{a}x + \frac{c}{a} \right ) = 0$

If the two roots are $\beta$ and $\lambda \beta$ then we know that
$\displaystyle a ( x - \beta ) (x - \lambda \beta ) = 0$

Expanding this and equating powers of x from your a, b, c equation form gives

$\displaystyle \frac{b}{a} = - \beta ( \lambda + 1 )$

and
$\displaystyle \frac{c}{a} = \lambda \beta ^2$

Solve the first of these for b and square it:
$\displaystyle b^2 = a^2 \beta ( \lambda + 1 )^2$

So
$\displaystyle \frac{b^2}{a} = a \beta ( \lambda + 1 )^2$

and since
$\displaystyle c = a \lambda \beta ^2$

we have, after some simplifying:
$\displaystyle \frac{b^2}{ac} = \frac{( \lambda + 1 )^2}{\lambda}$

You do the l, m, n quadratic the same way.

-Dan

**MichaelLight** · February 26th 2011, 09:08 PM

Okay... i solved part (i) but i face difficulty in solving (ii) now... please help...

**sa-ri-ga-ma** · February 26th 2011, 10:29 PM

According to the properties of quadratic equation, α + β = -p and αβ = q.

Then (α + β)^2 = .......

(α - β)^2 = (α + β)^2 - 4αβ = ........

(α^2 - β^2) = .........

**MichaelLight** · February 26th 2011, 11:20 PM

Originally Posted by sa-ri-ga-ma

According to the properties of quadratic equation, α + β = -p and αβ = q.

Then (α + β)^2 = .......

(α - β)^2 = (α + β)^2 - 4αβ = ........

(α^2 - β^2) = .........

I don't understand... why α + β = -p and αβ = q..? Isn't the given roots are (α^2 + β^2) and (α^2 - β^2)?

**sa-ri-ga-ma** · February 27th 2011, 08:47 AM

If α and β are the roots of ax^2 + bx + c = 0, then the sum of the roots = α + β = - b/a and product of the roots = αβ = c/a.

In the second equation,
(α^2 + β^2) + (α^2 - β^2) = -p and (α^2 + β^2) * (α^2 - β^2) = q.....(1)

Now (α^2 + β^2) = (α + β)^2 - 2αβ = ...?

(α - β)^2 = (α + β)^2 - 4αβ =.......? Find (α - β)

(α^2 - β^2) = (α + β)(α - β) = ......

Substitute these values in eq.(1)

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