Question

How do you solve \[\dfrac{{7x}}{{2x + 5}} + 1 = \dfrac{{10x - 3}}{{3x}}\]?

Answer 1

Hint: Here we need to solve for ‘x’. After taking LCM and cross multiplying we will have a quadratic equation. A polynomial of degree two is called a quadratic polynomial and its zeros can be found using many methods like factorization, completing the square, graphs, quadratic formula etc. The quadratic formula is used when we fail to find the factors of the equation.

Complete step-by-step solution:
Given,
\[\dfrac{{7x}}{{2x + 5}} + 1 = \dfrac{{10x - 3}}{{3x}}\]
Taking LCM on the left hand side of the equation we have,
\[\dfrac{{7x + 2x + 5}}{{2x + 5}} = \dfrac{{10x - 3}}{{3x}}\]
\[\dfrac{{9x + 5}}{{2x + 5}} = \dfrac{{10x - 3}}{{3x}}\]
Cross multiplying we have,
\[3x(9x + 5) = (2x + 5)(10x - 3)\].
\[27{x^2} + 15x = 2x(10x - 3) + 5(10x - 3)\]
\[27{x^2} + 15x = 20{x^2} - 6x + 50x - 15\]
\[27{x^2} + 15x - 20{x^2} + 6x - 50x + 15 = 0\]
\[27{x^2} - 20{x^2} + 15x + 6x - 50x + 15 = 0\]
\[7{x^2} - 29x + 15 = 0\]
On comparing the given equation with the standard quadratic equation \[a{x^2} + bx + c = 0\], we have \[a = 7\],\[b = - 29\] and \[c = 15\].
We cannot use the factorization method here, we are unable to split the middle term.
We have quadratic formula,
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Substituting we have,
\[ \Rightarrow x = \dfrac{{ - ( - 29) \pm \sqrt {{{( - 29)}^2} - 4(7)(15)} }}{{2(7)}}\]
\[ = \dfrac{{29 \pm \sqrt {841 - 420} }}{{14}}\]
\[ = \dfrac{{29 \pm \sqrt {421} }}{{14}}\]
Thus we have two roots,
\[ \Rightarrow x = \dfrac{{29 + \sqrt {421} }}{{14}}\] and \[x = = \dfrac{{29 - \sqrt {421} }}{{14}}\].
These are the solutions of \[\dfrac{{7x}}{{2x + 5}} + 1 = \dfrac{{10x - 3}}{{3x}}\].

Note: The highest exponent of the polynomial in a polynomial equation is called its degree. A polynomial equation has exactly as many roots as its degree. In various fields of mathematics require the point at which the value of a polynomial is zero, those values are called the factors or solution or zeros of the given polynomial. On the x-axis, the value of y is zero so the roots of an equation are the points on the x-axis that is the roots are simply the x-intercepts.