Sum of squares of roots

We have |2x-3| so the sign changes at x = 3/2

Let's take case 1: $$x\ge\dfrac{3}{2}$$

In this case: $$|2x-3| = 2x-3$$.

Hence, the equation will be $$x^{2} + (2x - 3) - 4 = 0$$

Or, $$x^{2} + 2x - 7 = 0$$

Solving using the discriminant formula i.e. $$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

$$x = \dfrac{-2 \pm \sqrt{2^2 - 4(1)(-7)}}{2(1)}$$

$$x = \dfrac{-2 \pm \sqrt{4 + 28}}{2}$$

$$x = \dfrac{-2 \pm \sqrt{32}}{2}$$

$$x = \dfrac{-2 \pm 4\sqrt{2}}{2}$$

$$x = -1 \pm 2\sqrt{2}$$

Now, let's check if the roots satisfy the initial assumption or not. 

$$x_1 = -1 + 2\sqrt{2} \approx -1 + 2(1.414) = 1.828$$. This is greater than 1.5, so it is a valid root.

$$x_2 = -1 - 2\sqrt{2} \approx -1 - 2.828 = -3.828$$. This is less than 1.5, so we reject it.

Let's take case 1:$$x<\dfrac{3}{2}$$

Here: $$|2x - 3| = -(2x - 3) = -2x + 3$$

Hence, the quadratic equation will be $$x^{2} + (-2x + 3) - 4 = 0$$

$$x^{2} - 2x - 1 = 0$$

Again, use the quadratic formula:

$$x = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)}$$

$$x = \dfrac{2 \pm \sqrt{4 + 4}}{2}$$

$$x = \dfrac{2 \pm \sqrt{8}}{2}$$

$$x = \dfrac{2 \pm 2\sqrt{2}}{2}$$

$$x = 1 \pm \sqrt{2}$$

Again, we need to check if the roots satisfy the range or not. 

$$x_3 = 1 + \sqrt{2} \approx 1 + 1.414 = 2.414$$. This is greater than $$1.5$$, so we reject it.

$$x_4 = 1 - \sqrt{2} \approx 1 - 1.414 = -0.414$$ This is less than $$1.5$$, so it is a valid root.

Now, calculating the sum of the squares:

$$\alpha = -1 + 2\sqrt{2}$$ and $$\beta = 1 - \sqrt{2}$$

$$\alpha^{2} = (-1 + 2\sqrt{2})^{2} = 1 - 4\sqrt{2} + 4(2) = 1 - 4\sqrt{2} + 8 = 9 - 4\sqrt{2}$$

$$\beta^{2} = (1 - \sqrt{2})^{2} = 1 - 2\sqrt{2} + 2 = 3 - 2\sqrt{2}$$

Adding them: $$\alpha^{2} + \beta^{2} = (9 - 4\sqrt{2}) + (3 - 2\sqrt{2})$$

$$\alpha^2+\beta^2=12-6\sqrt{2}\ =\ 6\left(2-\sqrt{2}\right)$$

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.

Given: $$3x^2 + \lambda x - 1 = 0$$ with roots $$\alpha$$ and $$\beta$$.

By Vieta's formulas:

$$ \alpha + \beta = -\frac{\lambda}{3}, \quad \alpha\beta = -\frac{1}{3} $$

We are given that the sum of the squares of the reciprocals of the roots is 15:

$$ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = 15 $$

$$ \frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} = 15 $$

Now, $$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \frac{\lambda^2}{9} - 2\left(-\frac{1}{3}\right) = \frac{\lambda^2}{9} + \frac{2}{3}$$

And $$\alpha^2\beta^2 = (\alpha\beta)^2 = \frac{1}{9}$$

Substituting:

$$ \frac{\frac{\lambda^2}{9} + \frac{2}{3}}{\frac{1}{9}} = 15 $$

$$ 9\left(\frac{\lambda^2}{9} + \frac{2}{3}\right) = 15 $$

$$ \lambda^2 + 6 = 15 $$

$$ \lambda^2 = 9 \implies \lambda = \pm 3 $$

Now we need to find $$6(\alpha^3 + \beta^3)^2$$.

Using the identity: $$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$$

$$ \alpha + \beta = -\frac{\lambda}{3} = \pm\frac{3}{3} = \pm 1 $$

$$ \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) $$

$$ = (\pm 1)^3 - 3\left(-\frac{1}{3}\right)(\pm 1) $$

$$ = \pm 1 + (\pm 1) = \pm 2 $$

Therefore:

$$ 6(\alpha^3 + \beta^3)^2 = 6 \times (\pm 2)^2 = 6 \times 4 = 24 $$

The answer is Option C: 24.