Question

How do you find the solution of the system of the equations $2p = q + 8$ and $q = 5\sqrt {p + 1} ?$

Answer 1

Hint: As we know that the above given set of equations are linear equations. An equation for a straight line is called a linear equation. The standard form of linear equations in two variables is $Ax + By = C$ . When an equation is given in this form it’s also pretty easy to find both intercepts $(x,y)$ . In this question we will solve by substituting the value of one variable into another.

Complete step by step solution:
As we know that the above given equation is a linear equation and to solve for $p$and $q$.
The first equation is $2p = q + 8$ and the second equation is $q = 5\sqrt {p + 1} $. We will put the value of $q$ in the first equation and we have: $2p = 5\sqrt {p + 1} + 8$.
We will transfer $8$ to the right hand side of the equation i.e. $2p - 8 = 5\sqrt {p + 1} $.
We will square both the sides of the equation : ${\left( {2p - 8} \right)^2} = {5^2}{(\sqrt {p + 1} )^2}$. By breaking the brackets and squaring both the sides we have, $4{p^2} - 32p + 64 = 25p + 25$.
By transferring the similar terms together we have $4{p^2} - 32p - 25p + 64 - 25 = 0 \Rightarrow 4{p^2} - 57p + 39 = 0$. We can see that it is a quadratic equation.
We will use the quadratic formula to solve it. The quadratic formula is $\dfrac{{ - b \pm \left( {{b^2} - 4ac} \right)}}{{2a}}$, where $b = - 57,a = 4$ and $c = 39$.
Now by applying the formula we have: $\dfrac{{ - ( - 57) \pm \sqrt {{{( - 57)}^2} - 4 \times 4 \times 39} }}{{2 \times 4}} = \dfrac{{57 \pm \sqrt {3249 - 624} }}{8}$.
On further solving we have $\dfrac{{57 \pm 51}}{8}$. It gives us two values i.e. $\dfrac{{57 - 51}}{8} = \dfrac{6}{8}or\dfrac{3}{4}$ and other value is $\dfrac{{57 + 51}}{8} = 13.5$.
By putting the values of $p$ in equation we have: $q = 5\sqrt {\dfrac{3}{4} + 1} = \dfrac{{5\sqrt 7 }}{2}$, while the other value is $q = 5\sqrt {13.5 + 1} = 5 \times \sqrt {14.5} $. By approximating the value of square root $q = 5 \times 3.80 = 19$(approx.)
Hence these are the required solutions of the quadratic equation.

Note: The formula that we have used above is called the difference of squares formula i.e.${(a - b)^2} = {a^2} - 2ab + {b^2}$. We have also approximated the values of the square roots. We should keep in mind the positive and negative signs while calculating the value of any variable as it will change it’s slope and value. In the equation $Ax + By = C$ ,$A$ and $B$ are real numbers and $C$ is a constant, it can be equal to zero $(0)$.