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Question: How to solve by quadratic formula: \( (p - q)x + \dfrac{{2q}}{x} = (p + q)? \)...

How to solve by quadratic formula: (p−q)x+2qx=(p+q)?(p - q)x + \dfrac{{2q}}{x} = (p + q)?

Explanation

Solution

First of all we will convert the given equation in the form of a standard quadratic equation then will compare the simplified equation with it and will find out the roots using the formulas.

Complete step-by-step solution:
Take the given equation: (p−q)x+2qx=(p+q)(p - q)x + \dfrac{{2q}}{x} = (p + q)
Take L.C.M. (least common multiple) on the left hand side of the above equation.
(p−q)x2+2qx=(p+q)\dfrac{{(p - q){x^2} + 2q}}{x} = (p + q)
Do cross multiplication where the denominator of one side is multiplied with the numerator of the opposite side.
(p−q)x2+2q=(p+q)x(p - q){x^2} + 2q = (p + q)x
Now bring all the terms on the left hand side of the equation. When you move any term from one side to another, then the sign of the term also changes. Positive terms become negative and vice-versa.
(p−q)x2+2q−(p+q)x=0(p - q){x^2} + 2q - (p + q)x = 0
Rearranging the above equation.
(p−q)x2−(p+q)x+2q=0(p - q){x^2} - (p + q)x + 2q = 0
Let us consider the general form of the quadratic equation.
ax2+bx+c=0a{x^2} + bx + c = 0
Compare the standard equation with the simplified equation (p−q)x2−(p+q)x+2q=0(p - q){x^2} - (p + q)x + 2q = 0 .....(A)
⇒a=(p−q) ⇒b=−(p+q) ⇒c=2q  \Rightarrow a = (p - q) \\\ \Rightarrow b = - (p + q) \\\ \Rightarrow c = 2q \\\
Now, Δ=b2−4ac\Delta = {b^2} - 4ac
Place the values in the above equation.
Δ=(−(p+q))2−4(p−q)2q\Delta = {( - (p + q))^2} - 4(p - q)2q
Square of negative terms also gives the resultant value as the positive term. Simplify the above equation.
Δ=p2+2pq+q2−8q(p−q)\Delta = {p^2} + 2pq + {q^2} - 8q(p - q)
When there is a negative sign outside the bracket then the sign of the terms inside the bracket changes. Positive term becomes negative and negative term becomes positive.
Δ=p2+2pq+q2−8qp+8q2\Delta = {p^2} + 2pq + {q^2} - 8qp + 8{q^2}
Simplify between the like terms.
Δ=p2−6pq+9q2\Delta = {p^2} - 6pq + 9{q^2}
This can be written as the whole square of the difference of the two terms.
Δ=(p−3q)2\Delta = {(p - 3q)^2}
Take square-root on both the sides of the above equation.
Δ=(p−3q)2\sqrt \Delta = \sqrt {{{(p - 3q)}^2}}
Square and square root cancel each other.
Δ=(p−3q)\sqrt \Delta = (p - 3q)
Now, the roots of the equation can be given by
x=−b±Δ2ax = \dfrac{{ - b \pm \sqrt \Delta }}{{2a}}
Place the values in the above equation-
x=(p+q)±(p−3q)2(p−q)x = \dfrac{{(p + q) \pm (p - 3q)}}{{2(p - q)}}
So the roots will be
x=(p+q)+(p−3q)2(p−q)x = \dfrac{{(p + q) + (p - 3q)}}{{2(p - q)}} or x=(p+q)−(p−3q)2(p−q)x = \dfrac{{(p + q) - (p - 3q)}}{{2(p - q)}}
Simplify the above equation.
x=2p−2q2(p−q)x = \dfrac{{2p - 2q}}{{2(p - q)}} or x=(p+q−p+3q)2(p−q)x = \dfrac{{(p + q - p + 3q)}}{{2(p - q)}}
Common factors from the numerator and the denominator cancel each other and like terms with the same value and opposite sign cancel each other.
⇒x=1\Rightarrow x = 1 or x=4q2(p−q)x = \dfrac{{4q}}{{2(p - q)}}
Hence, the required solution is –
x=1x = 1 or x=2q(p−q)x = \dfrac{{2q}}{{(p - q)}}

Note: Be careful about the sign convention, when you open the brackets. When there is a negative sign outside the bracket then the sign of all the terms inside the bracket changes. Positive terms become negative and vice-versa. Remember the standard form of the quadratic equation and delta formula for the efficient and an accurate solution.