Quadratic formula (roots)

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 systematically, let's break down the Right Hand Side (RHS) first.

The function $$\min\{|x - 3|, |x + 2|\}$$ gives the smaller of the two distances from $$x$$ to $$3$$ and from $$x$$ to $$-2$$. The point where these distances are exactly equal is the midpoint:

$$x = \frac{3 + (-2)}{2} = 0.5$$

This allows us to redefine the RHS piecewise:

• For $$x \le 0.5$$, the distance to $$-2$$ is smaller, so $$\text{RHS} = |x + 2|$$
• For $$x > 0.5$$, the distance to $$3$$ is smaller, so $$\text{RHS} = |x - 3|$$

Now, split the original equation into these two main regions and solve.

Region 1: For $$x \le 0.5$$

The equation becomes: $$x^2 + 3x + 2 = |x + 2|$$

Since the absolute value changes behavior at $$x = -2$$, we check two sub-cases:

• Case 1a ($$x \le -2$$):

  $$x^2 + 3x + 2 = -(x + 2)$$

  $$x^2 + 4x + 4 = 0$$

  $$(x + 2)^2 = 0 \implies x = -2$$

  (Since $$-2 \le -2$$, this is a valid solution).

• Case 1b ($$-2 < x \le 0.5$$):

  $$x^2 + 3x + 2 = +(x + 2)$$

  $$x^2 + 2x = 0$$

  $$x(x + 2) = 0 \implies x = 0 \text{ or } x = -2$$

  (Since $$x = 0$$ falls within $$-2 < x \le 0.5$$, this is a valid solution).

Region 2: For $$x > 0.5$$

The equation becomes: $$x^2 + 3x + 2 = |x - 3|$$

Again, split at the critical point $$x = 3$$:

• Case 2a ($$0.5 < x < 3$$):

  $$x^2 + 3x + 2 = -(x - 3)$$

  $$x^2 + 4x - 1 = 0$$

  Using the quadratic formula: $$x = \frac{-4 \pm \sqrt{16 - 4(1)(-1)}}{2} = -2 \pm \sqrt{5}$$

  Since $$\sqrt{5} \approx 2.23$$, the roots are roughly $$0.23$$ and $$-4.23$$. Neither of these falls in the interval $$(0.5, 3)$$.

  (No solution here).

• Case 2b ($$x \ge 3$$):

  $$x^2 + 3x + 2 = +(x - 3)$$

  $$x^2 + 2x + 5 = 0$$

  Checking the discriminant: $$D = 2^2 - 4(1)(5) = -16$$

  Since $$D < 0$$, there are no real roots.

  (No solution here).

Conclusion:

The only valid real solutions are $$x = -2$$ and $$x = 0$$.

Therefore, the number of real solutions is 2.

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

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