Question

A girl goes to school at x km/hr. and returns at (x - 4) km/hr. The product of both her speeds is - 4, find the value of x.
(a) 1
(b) 2
(c) -2
(d) 3

Answer 1

To find the value of $x$, first, you need to create an equation according to the condition in the question and solve the equation by finding its roots which in turn is the value of $x$. Use the factorization method to solve the quadratic equation.

Complete step-by-step solution:
Here, we have been given that, the girl goes to school at $x$ km/hr. and returns from school at $\left( x-4 \right)$ km/hr.
Girl goes to school at speed km/hr. = $x$
Girl returns from school at a speed km/hr. = $\left( x-4 \right)$
Product of both the speeds = - 4
According to the given data,
$x\left( x-4 \right)=-4$
Now further, solving this equation we will get the value of $x$.
First open the brackets by multiplying $x$ throughout,
$\begin{aligned} & {{x}^{2}}-4x=-4 \\\ &\Rightarrow {{x}^{2}}-4x+4=0 \end{aligned}$
Now, the equation we got is quadratic equation, whose general form is $a{{x}^{2}}+bx+c=0$
Let us find the roots to find the value of the $x$.
$\begin{aligned} & {{x}^{2}}-2x-2x+4=0 \\\ &\Rightarrow x\left( x-2 \right)-2\left( x-2 \right)=0 \\\ &\Rightarrow \left( x-2 \right)\left( x-2 \right)=0 \end{aligned}$
Here we have, ${{\left( x-2 \right)}^{2}}=0$
After taking the square root on both the sides,
$\left( x-2 \right)$= 0
$\Rightarrow x=2$
Hence, the value of $x=2$, it is also the speed at which the girl went to her school.
So, the girl went to her school at a speed of 2 km/hr.

Note: When you are solving a quadratic equation, always select your roots carefully whilst following the rules of factorization. Also do not forget to put the units for the required values. You can also find the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ by the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, which also gives the value of $x$.