MHB Louis's Question from YahooAnswers:Fp1 Polynomial and roots question Help?

[CaptainBlack]

Question:
1.Find the range of values of \(a\) for which \[(2-3a)x^2+(4-a)x+2=0\]has real roots.2. If the roots of the equation \(4x^3+7x^2-5x-1=0\) are \(\alpha\) , \(\beta\) and \( \gamma\),find the equation whose roots are:
(a) \( \alpha+1,\beta+1\) and \(\gamma+1\)
(b) \(\alpha^2 \beta^2\) and \( \gamma^2 \)CB

 

[CaptainBlack]

Answer:
Part 1.
For a quadratic \(ax^2+bx+c\) the discriminant is \(b^2-4ac\) this is the term that appears under the square root sign in the quadratic formula: \[x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}\]for the solution of \(ax^2+bx+c=0\). The quadratic equation has real roots precisely when the discriminant is non-negative.The discriminant of \((2-3a)x^2+(4-a)x+2\) is \((4-a)^2-4(2-3a)2=a^2+16a\). Because the discriminant is positive for large \(|a|\) the discriminant is negative between the roots of \(a^2+16a\) which is on the interval \( (-16,0)\). Hence the discriminant is non-negative when \(a \lt -16\) or \(a \gt 0\) and so \((2-3a)x^2+(4-a)x+2=0\) has real roots when \(a \lt -16\) or \(a \gt0\).Part 2.
You may know Vieta's relations for the roots and coefficients of polynomials or cubics in particular or we can derive them:For a cubic \(ax^3+bx^2+cx+d\) with roots \(\alpha,\beta\) and \(\beta\) we have:\[\begin{aligned}a(x-\alpha)(x-\beta)(x-\gamma)&=ax^3-a(\alpha+\beta+\gamma)x^2+a(\alpha\beta+\alpha \gamma+\beta \gamma)x-a(\alpha\beta \gamma)\\ &=ax^3+bx^2+cx+d \end{aligned}\] means that \(b=-a(\alpha+\beta+\gamma)\), \(c=a(\alpha\beta+\alpha \gamma+\beta \gamma)\) and \(d=-a(\alpha\beta \gamma)\).So for the cubic \(4x^3+7x^2-5x-1\) we have:
\(\alpha+\beta+\gamma=-7/4\)
\(\alpha\beta+\alpha\gamma+\beta\gamma=-5/4\)
\(\alpha\beta\gamma=1/4\)(a) Let the equation with roots \( \alpha+1,\beta+1\) and \(\gamma+1\) be:\[x^3+ux^2+vx+w=0\] (we can of course multiply this by any constant we want if we want the left hand side to not be monic). Now from the relations between coefficients and roots we know that \[\begin{aligned}w&=-( \alpha+1)(\beta+1)(\gamma+1)\\ &=-(\alpha \beta \gamma + \alpha\beta + \alpha \gamma + \beta \gamma +\alpha+\beta+\gamma +1) \\ &= - \frac{1}{4} + \frac{5}{4} + \frac{7}{4} -1=\frac{7}{4} \end{aligned}\]
\[\begin{aligned}v&=( \alpha+1)(\beta+1)+(\alpha+1)(\gamma+1)+(\beta+1)(\gamma+1) \\
&=\alpha\beta+\alpha+\beta+1+\alpha \gamma+\alpha+\gamma+1+\beta \gamma+\beta+\gamma +1\\
&=(\alpha\beta+\alpha \gamma+\beta \gamma)+2(\alpha+\beta+\gamma)+3\\
&=\dots
\end{aligned}\]...to be continued? Probably not, there is enough here to indicate how to complete the solution.

CB

 

[MarkFL]

For part a) of question 2, we could simply translate the polynomial 1 unit to the right, i.e.:

$\displaystyle 4(x-1)^3+7(x-1)^2-5(x-1)-1=0$

 

[CaptainBlack]

For part a) of question 2, we could simply translate the polynomial 1 unit to the right, i.e.:

$\displaystyle 4(x-1)^3+7(x-1)^2-5(x-1)-1=0$

Yes, but since we are going to need Viete's relations for the next part (and I don't want to provide two different approaches we might as well use them for both parts)

cb