Symmetric Expressions in Roots

Given $$x^2 + 2\sqrt{2}x - 1 = 0$$ with roots $$\alpha, \beta$$.

By Vieta's: $$\alpha + \beta = -2\sqrt{2}$$, $$\alpha\beta = -1$$.

$$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 8 + 2 = 10$$

$$\alpha^4 + \beta^4 = (\alpha^2+\beta^2)^2 - 2(\alpha\beta)^2 = 100 - 2 = 98$$

$$\alpha^6 + \beta^6 = (\alpha^2+\beta^2)^3 - 3(\alpha\beta)^2(\alpha^2+\beta^2) = 1000 - 3(1)(10) = 970$$

So the two roots of the required equation are: $$\alpha^4 + \beta^4 = 98$$ and $$\frac{1}{10}(\alpha^6+\beta^6) = \frac{970}{10} = 97$$.

Sum of roots = $$98 + 97 = 195$$.

Product of roots = $$98 \times 97 = 9506$$.

The equation is $$x^2 - 195x + 9506 = 0$$.

The correct answer is Option 3.

We are given $$p + q = 2$$ and $$p^4 + q^4 = 272$$, and we need to find the quadratic equation whose roots are $$p$$ and $$q$$.

We know that $$p^2 + q^2 = (p + q)^2 - 2pq = 4 - 2pq$$.

Now, $$p^4 + q^4 = (p^2 + q^2)^2 - 2p^2q^2$$. Substituting, we get $$(4 - 2pq)^2 - 2(pq)^2 = 272$$.

Let $$s = pq$$. Then $$16 - 16s + 4s^2 - 2s^2 = 272$$, which gives $$2s^2 - 16s + 16 = 272$$.

So $$2s^2 - 16s - 256 = 0$$, or $$s^2 - 8s - 128 = 0$$.

Using the quadratic formula, $$s = \frac{8 \pm \sqrt{64 + 512}}{2} = \frac{8 \pm \sqrt{576}}{2} = \frac{8 \pm 24}{2}$$.

This gives $$s = 16$$ or $$s = -8$$.

The quadratic equation with sum of roots $$p + q = 2$$ and product $$pq = 16$$ is $$x^2 - 2x + 16 = 0$$, and with $$pq = -8$$ it is $$x^2 - 2x - 8 = 0$$. Among the given options, only $$x^2 - 2x + 16 = 0$$ is present.

Hence, the correct answer is Option D.

We start with the quadratic equation $$7x^2-3x-2=0$$ whose roots are denoted by $$\alpha$$ and $$\beta$$. For any quadratic equation $$ax^2+bx+c=0$$ the standard Vieta’s relations tell us

$$\alpha+\beta=-\dfrac{b}{a}, \qquad \alpha\beta=\dfrac{c}{a}.$$

Here we have $$a=7,\; b=-3,\; c=-2,$$ so

$$\alpha+\beta=-\dfrac{-3}{7}=\dfrac{3}{7}, \qquad \alpha\beta=\dfrac{-2}{7}.$$

We are asked to find the value of

$$S=\dfrac{\alpha}{1-\alpha^2}+\dfrac{\beta}{1-\beta^2}.$$

To simplify each denominator, we use the fact that every root satisfies the original quadratic. Because $$7\alpha^2-3\alpha-2=0,$$ we can solve for $$\alpha^2$$:

$$7\alpha^2=3\alpha+2\;\; \Longrightarrow\;\; \alpha^2=\dfrac{3\alpha+2}{7}.$$

Subtracting this from 1 gives

$$1-\alpha^2=1-\dfrac{3\alpha+2}{7}=\dfrac{7-(3\alpha+2)}{7}=\dfrac{5-3\alpha}{7}.$$

Hence

$$\dfrac{\alpha}{1-\alpha^2}=\dfrac{\alpha}{\dfrac{5-3\alpha}{7}} =\dfrac{7\alpha}{5-3\alpha}.$$

Exactly the same reasoning for $$\beta$$ gives

$$\dfrac{\beta}{1-\beta^2}=\dfrac{7\beta}{5-3\beta}.$$

Therefore

$$S=\dfrac{7\alpha}{5-3\alpha}+\dfrac{7\beta}{5-3\beta} =7\left(\dfrac{\alpha}{5-3\alpha}+\dfrac{\beta}{5-3\beta}\right).$$

We now combine the two fractions inside the parentheses by taking a common denominator:

$$\dfrac{\alpha}{5-3\alpha}+\dfrac{\beta}{5-3\beta} =\dfrac{\alpha(5-3\beta)+\beta(5-3\alpha)} {(5-3\alpha)(5-3\beta)}.$$

Expanding the numerator carefully, we obtain

$$\alpha(5-3\beta)+\beta(5-3\alpha) =5\alpha-3\alpha\beta+5\beta-3\alpha\beta =5(\alpha+\beta)-6\alpha\beta.$$

The denominator multiplies out to

$$(5-3\alpha)(5-3\beta)=25-15(\alpha+\beta)+9\alpha\beta.$$

We now substitute the known values $$\alpha+\beta=\dfrac{3}{7}$$ and $$\alpha\beta=-\dfrac{2}{7}$$ into both numerator and denominator.

Numerator:

$$5(\alpha+\beta)-6\alpha\beta =5\left(\dfrac{3}{7}\right)-6\left(-\dfrac{2}{7}\right) =\dfrac{15}{7}+\dfrac{12}{7} =\dfrac{27}{7}.$$

Denominator:

$$25-15(\alpha+\beta)+9\alpha\beta =25-15\left(\dfrac{3}{7}\right)+9\left(-\dfrac{2}{7}\right) =25-\dfrac{45}{7}-\dfrac{18}{7} =25-\dfrac{63}{7} =25-9 =16.$$

So the inner fraction becomes

$$\dfrac{\dfrac{27}{7}}{16} =\dfrac{27}{7\times16}.$$

Finally we multiply by the outer factor of 7:

$$S=7\left(\dfrac{27}{7\times16}\right)=\dfrac{27}{16}.$$

Hence, the correct answer is Option D.