# A girl goes to her friend's house, which is at a distance of 12 km. She covers half of the distance at a speed of x km/hr. and the remaining distance at a speed of (x+2)km/hr. If she takes 2 hrs 30 minutes to cover the whole distance, find "x" ?

$4$.

#### Explanation:

This is simple equation problem.

The velocity for half the distance = $x$ km/h.

So, Time taken to cross the first half = $\frac{\frac{12}{2}}{x}$ hours.

The velocity for the other half = $x + 2$ km/h.

So, Time taken to cross the rest = $\frac{\frac{12}{2}}{x + 2}$ hours.

And, $2$ hours and $30$ minutes = $2 \frac{1}{2}$ hours.

So, On with the equation.

$\textcolor{w h i t e}{\times x} \frac{\frac{12}{2}}{x} + \frac{\frac{12}{2}}{x + 2} = 2 \frac{1}{2}$

$\Rightarrow \frac{6}{x} + \frac{6}{x + 2} = \frac{5}{2}$

$\Rightarrow 6 \left(\frac{1}{x} + \frac{1}{x + 2}\right) = \frac{5}{2}$ [Taking the 6 as common part]

$\Rightarrow \frac{\left(x + 2\right) + x}{x \left(x + 2\right)} = \frac{5}{2} \cdot \frac{1}{6}$ [Transposed the 6]

$\Rightarrow \frac{x + 2 + x}{{x}^{2} + 2 x} = \frac{5}{12}$

$\Rightarrow \frac{2 x + 2}{{x}^{2} + 2 x} = \frac{5}{12}$ [Simplified using L.C.M]

$\Rightarrow 24 x + 24 = 5 {x}^{2} + 10 x$ [Cross multiplication]

$\Rightarrow 5 {x}^{2} + 10 x - 24 x - 24 = 0$ [Transposing everything to a side of the equation]

$\Rightarrow 5 {x}^{2} - 14 x - 24 = 0$

Now, Use The Sridhar Acharya's Rule or Quadratic Formula to solve this.

And, Don't forget to check the Discriminant first.

$D = {b}^{2} - 4 a c = {\left(- 14\right)}^{2} - 4 \cdot 5 \cdot \left(- 24\right) = 196 + 480 = 676 > 0$

So, Definitely There are two real solutions for $x$ and both are distinct.

Now first solution for $x$ = $\frac{- b + \sqrt{D}}{2 a} = \frac{- \left(- 14\right) + \sqrt{676}}{2 \cdot 5}$

$\textcolor{w h i t e}{\times \times \times \times \times \times \times} = \frac{14 + 26}{10} = \frac{40}{10} = 4$

Now, The second solution for $x$

= $\frac{- b - \sqrt{D}}{2 a} = \frac{- \left(- 14\right) - \sqrt{676}}{2 \cdot 5}$

$\textcolor{w h i t e}{\times \times \times \times \times \times \times} = \frac{14 - 26}{10} = - \frac{12}{10} = - 1.2$

So, $x = 4 , - 1.2$.

But speed can't be negative in any circumstances as distance travelled can't be negative.

So, $x = 4$ is the legitimate value.