Quadratic Equation

Let the time taken by A to complete the construction be $$ a$$ days. Then B’s time = $$ a + 24$$ days.

If the total work is say 1 unit, then work done by A in one day is $$\frac{1}{a} $$ units , and work by B is $$ \frac{1}{a+24} $$ units.

Given, $$  \dfrac{1}{a} + \dfrac{1}{a+24} = \dfrac{1}{22.5} $$ [work done by A in one day+ work by B in one day= total work done in one day]

or, $$  \dfrac{(a+24) + a}{a(a+24)} = \dfrac{1}{22.5} $$

or, $$  \dfrac{2a + 24}{a(a+24)} = \dfrac{1}{22.5} $$

or, $$ 22.5(2a + 24) = a(a+24) $$

or, $$ 45a + 540 = a^2 + 24a$$ 

or, $$ a^2 - 21a - 540 = 0 $$

or, $$ a = \dfrac{21 \pm \sqrt{21^2 + 4 \cdot 540}}{2} $$

or, $$ a = \dfrac{21 \pm \sqrt{441 + 2160}}{2}= \dfrac{21 \pm \sqrt{2601}}{2} = \dfrac{21 \pm 51}{2} $$

Since, the number of days cannot be negative, so $$ a = \dfrac{72}{2} = 36 \text{days} $$

To solve this equation, let's look for a way to simplify it using substitution.

Notice that the terms $$(x - 6\sqrt{x})$$ and the $$9$$ from the last constant term can be grouped to form a perfect square:

$$x - 6\sqrt{x} + 9 = (\sqrt{x} - 3)^2 = |\sqrt{x} - 3|^2$$

Substitute this back into the original equation to reorganize it:

$$(-2 + \sqrt{3})|\sqrt{x} - 3| + |\sqrt{x} - 3|^2 - 2\sqrt{3} = 0$$

Let $$t = |\sqrt{x} - 3|$$. Since $$t$$ represents an absolute value, we know that $$t \ge 0$$.

The equation now becomes a standard quadratic in terms of $$t$$:

$$t^2 + (-2 + \sqrt{3})t - 2\sqrt{3} = 0$$

We can factor this quadratic by splitting the middle term:

$$t^2 - 2t + \sqrt{3}t - 2\sqrt{3} = 0$$

$$t(t - 2) + \sqrt{3}(t - 2) = 0$$

$$(t - 2)(t + \sqrt{3}) = 0$$

This gives two possible values for $$t$$:

$$t = 2$$ or $$t = -\sqrt{3}$$

Since we established that $$t \ge 0$$, we reject $$-\sqrt{3}$$. Therefore, $$t = 2$$.

Substitute back $$t = |\sqrt{x} - 3|$$:

$$|\sqrt{x} - 3| = 2$$

This results in two distinct cases:

1. $$\sqrt{x} - 3 = 2 \implies \sqrt{x} = 5 \implies x = 25$$
2. $$\sqrt{x} - 3 = -2 \implies \sqrt{x} = 1 \implies x = 1$$

The roots of the equation are $$1$$ and $$25$$. We are given that $$\alpha$$ and $$\beta$$ are the roots with $$\alpha < \beta$$, which means:

• $$\alpha = 1$$
• $$\beta = 25$$

Finally, substitute these values into the expression we need to evaluate:

$$\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta}$$

$$= \sqrt{\frac{25}{1}} + \sqrt{(1)(25)}$$

$$= 5 + 5 = 10$$

The equation given to us is $$x\lvert x+4 \rvert + 3\lvert x+2 \rvert + 10 = 0$$

Critical points for this are $$ x = -4, -2$$

Case 1: $$ x \ge -2$$

Then, $$ |x+4| = x+4,\quad |x+2| = x+2 $$

The equation becomes, 

$$ x(x+4) + 3(x+2) + 10 = 0$$
or, $$ x^2 + 4x + 3x + 6 + 10 = 0$$
or, $$  x^2 + 7x + 16 = 0 $$

For this equation, the discriminant is 49 - 64 = -15 < 0. So, this has no real solution.

Case 2: $$ -4 \le x < -2$$

Then, $$ |x+4| = x+4,\quad |x+2| = -(x+2) $$

The equation becomes,

$$ x(x+4) + 3(-(x+2)) + 10 = 0$$
or, $$ x^2 + 4x - 3x - 6 + 10 = 0 $$
or, $$ x^2 + x + 4 = 0 $$

For this equation too, the discriminant is  1 - 16 = -15 < 0. So this too has no real solution.

Case 3: $$ x < -4$$

Then, $$ |x+4| = -(x+4),\quad |x+2| = -(x+2) $$

The equation becomes, 

$$x(-(x+4)) + 3(-(x+2)) + 10 = 0$$
or, $$ x^2 - 4x - 3x - 6 + 10 = 0 $$
or, $$ x^2 - 7x + 4 = 0 $$

or, $$ x = \dfrac{-7 \pm \sqrt{49 + 16}}{2} = \dfrac{-7 \pm \sqrt{65}}{2} $$

Now we check if the solution values lie in the domain x < -4:

• $$ \frac{-7 + \sqrt{65}}{2} \approx 0.53$$ , which is not a valid solution
• $$ \frac{-7 - \sqrt{65}}{2} \approx -7.53$$ , which is a valid solution. 

Hence, the number of distinct real solutions is 1.