Question

Solve the following quadratic equation using factorization method:
\[\dfrac{2}{{x + 1}} + \dfrac{3}{{2(x - 2)}} = \dfrac{{23}}{{5x}}\], where $x \ne 0, - 1,2$
State true or false, whether the root of the given quadratic equation is $4$.

Answer 1

Hint: For solving any quadratic equation of the form $a{x^2} + bx + c = 0$ using the factorization method, then our main objective is to write it in the form of $(x - \alpha )(x - \beta ) = 0$ where $(x - \alpha )$ and $(x - \beta )$ are the factors and $\alpha $and $\beta $are the roots of the given quadratic equation.
 To do so need to split the term $bx$ in ${b_1}x + {b_2}x$ such that the sum of ${b_1}$ and ${b_2}$ is $b$ and the product of ${b_1}$ and ${b_2}$ is $a \times c$.

Complete step-by-step solution:
The given quadratic equation is \[\dfrac{2}{{x + 1}} + \dfrac{3}{{2(x - 2)}} = \dfrac{{23}}{{5x}}\].
Solve the left side of the equation to make it into the standard form of the equation.
\[\dfrac{{4(x - 2) + 3(x + 1)}}{{2(x - 2)(x + 1)}} = \dfrac{{23}}{{5x}}\]
\[\dfrac{{7x - 5}}{{2({x^2} - x - 2)}} = \dfrac{{23}}{{5x}}\]
Cross multiply the above equation,
\[(7x - 5)5x = 46({x^2} - x - 2)\]
\[35{x^2} - 25x = 46{x^2} - 46x - 92\]
\[11{x^2} - 21x - 92 = 0\]
Compare the given equation with the standard form of equation $a{x^2} + bx + c = 0$.
We get, $a = 11,b = - 21,c = - 92$
Now, we need to split the term $by$ in ${b_1}y + {b_2}y$ such that the sum of ${b_1}$ and ${b_2}$ is $b$ and the product of ${b_1}$ and ${b_2}$ is $a \times c$.
Using the given terms, we get
\[{b_1} + {b_2} = - 21\]
\[{b_1} \times {b_2} = 11 \times ( - 92) = 1012\]
Let us take \[{b_1} = 23\] and \[{b_2} = - 44\], since it satisfies both the above conditions,
\[{b_1} + {b_2} = (23) + ( - 44) = - 21\]
\[{b_1} \times {b_2} = (23) \times ( - 44) = - 1012\]
We split $ - 21x$ term into $23x$ and $ - 44x$,
\[11{x^2} + 23x - 44x - 92 = 0\]
Take $x$ common from the terms $11{x^2} + 23x$ and take $ - 4$ common from the terms $ - 44x - 92$, to obtain
\[x(11x + 23) - 4(11x + 23) = 0\]
Take the term $(11x + 23)$common from the above equation, we get
$(11x + 23)(x - 4) = 0$
The above equation can further be written as,
$(11x + 23) = 0$ or $(x - 4) = 0$
$x = - \dfrac{{23}}{{11}}$ or $x = 4$
So, the roots of the quadratic equation \[\dfrac{2}{{x + 1}} + \dfrac{3}{{2(x - 2)}} = \dfrac{{23}}{{5x}}\] are $ - \dfrac{{23}}{{11}}$ and $4$. So, the statement that the given quadratic equation has the root $4$ is true.

Note: A quadratic equation is a quadratic polynomial of degree $2$, which is equated to $0$ and thus we get the quadratic equation of the form $a{x^2} + bx + c = 0$, where $a,b$ and $c$ belong to the set of real numbers and $a$ cannot be zero. The roots of any quadratic equation are the set of real numbers that satisfy the given equation.