Polynomial Equation - Roots Formulas

Let the roots of the given equation $$5x^{3} + cx^{2} - 10x + 9 = 0$$ be r, -r and p

r - r + p = $$-\frac{c}{5}$$

p = $$-\frac{c}{5}$$ ...... (1)

$$-r^2-pr+pr=-2$$

$$r^2=2$$ ...... (2)

$$-r^2p=-\frac{9}{5}$$

$$p=\frac{9}{10}$$ ...... (3)

Substituting p in (1), we get

$$\frac{9}{10}=-\frac{c}{5}$$

$$-\frac{9}{2}=c$$

The answer is option A.

Given a and b are the distinct roots of the equation $$2x^{2} - 6x + k = 0$$

=> a + b = -(-6/2) = 3 (Sum of the roots)

=> ab = k/2 (Product of the roots)

Now, (a+b) and ab are the roots of the quadratic equation $$x^{2} + px + p = 0$$

=> a + b + ab = -p => 3 + k/2 = -p ---(1)

=> (a + b)(ab) = p => 3(k/2) = p ---(2)

$$3+\dfrac{k}{2}=-\dfrac{3k}{2}$$ => 2k = -3 => k = $$-\dfrac{3}{2}$$

p = $$\dfrac{3k}{2}=\dfrac{3}{2}\left(-\dfrac{3}{2}\right)=-\dfrac{9}{4}$$

=> 8(k-p) = $$8\left(-\frac{3}{2}+\frac{9}{4}\right)=-12+18=6$$

It is given that $$x^2 + bx + c = 0$$ has two real roots. Let the roots of the equation be $$\alpha\ ,\beta\ $$. ($$\alpha\ \ >\ \beta\ $$)

Then, we can say that $$\frac{1}{\alpha\ }-\frac{1}{\beta\ }=\frac{1}{3}$$ .... Eq(1)

Similarly, $$\frac{1}{\alpha\ ^2}+\frac{1}{\beta^2\ }=\frac{5}{9}$$ .... Eq (2)

Eq(2) can be written as $$\left(\frac{1}{\alpha\ }-\frac{1}{\beta\ }\right)^{^2}+2\cdot\frac{1}{\alpha\ }\cdot\frac{1}{\beta\ }=\frac{5}{9}$$

=> $$\left(\frac{1}{3}\right)^{^2}+2\cdot\frac{1}{\alpha\ }\cdot\frac{1}{\beta\ }=\frac{5}{9}$$

=> $$\frac{2}{\alpha\ \cdot\beta\ }=\frac{4}{9}=>\ \frac{1}{\alpha\ \cdot\beta\ }=\frac{2}{9}$$

=> $$\alpha\ \cdot\beta\ =\frac{9}{2}$$

We know that the product of the roots is equal to c, which implies$$c=\frac{9}{2}$$

We also know that the sum of the roots is equal to -b. 

=> $$\frac{1}{\alpha\ ^2}+\frac{1}{\beta\ ^2}=\left(\frac{1}{\alpha\ }+\frac{1}{\beta\ }\right)^{^2}-\frac{2}{\alpha\ \beta\ }=\frac{5}{9}$$

=> $$\left(\ \frac{\ \alpha\ +\beta\ }{\alpha\ \beta\ }\right)^{^2}-\frac{4}{9}=\frac{5}{9}$$

=> $$\left(\ \frac{\ \alpha\ +\beta\ }{\alpha\ \beta\ }\right)^{^2}=\left(1\right)^2$$

=> $$\alpha\ +\beta\ \ =\ \pm\ \alpha\ \beta\ $$

Hence, the maximum value of b is $$\frac{9}{2}$$.

Hence, the maximum value of (b+c) is 9