Question

Solve the following quadratic equation by factorization
\[\dfrac{x-1}{x-2}+\dfrac{x-3}{x-4}=3\dfrac{1}{3}\left( x\ne 2,4 \right)\]

Answer 1

Hint: We multiply $\left( x-2 \right)\left( x-4 \right)$ with all the terms on both sides of the equation. We simplify until we get a quadratic equation of the form $a{{x}^{2}}+bx+c=0$ and use the splitting the middle term method to factorize it where we have to find the two factors $p,q$ such that $p+q=b$ and $pq=a\times c$.\[\]

Complete step-by-step solution:
We know that the general form of the quadratic equation is $a{{x}^{2}}+bx+c=0$ where $a,b,c$ are real numbers with the condition $a\ne 0$. The given quadratic equation from the question is
\[\dfrac{x-1}{x-2}+\dfrac{x-3}{x-4}=3\dfrac{1}{3}\left( x\ne 2,4 \right)\]
The above quadratic equation is not in the general form and is given in linear polynomial fractions. The polynomial fractions exists because $x-2\ne 0,x-4\ne 0$ since we are given the condition $x\ne 2,4$. Let us multiply $\left( x-2 \right)\left( x-4 \right)$ with all the terms on both sides of the equation. We have
\[\begin{align}
  & \dfrac{x-1}{x-2}\left( x-2 \right)\left( x-4 \right)+\dfrac{x-3}{x-4}\left( x-2 \right)\left( x-4 \right)=3\dfrac{1}{3}\left( x-2 \right)\left( x-4 \right) \\
 & \Rightarrow \left( x-1 \right)\left( x-4 \right)+\left( x-3 \right)\left( x-2 \right)=\dfrac{10}{3}\left( x-2 \right)\left( x-4 \right) \\
\end{align}\]
Let us multiply the linear polynomial factors and have
\[\begin{align}
  & \Rightarrow {{x}^{2}}-4x+4+{{x}^{2}}-6x+6=\dfrac{10}{3}\left( {{x}^{2}}-6x+8 \right) \\
 & \Rightarrow 2{{x}^{2}}-10x+10=\dfrac{10{{x}^{2}}-60x+80}{3} \\
\end{align}\]
Let us cross multiply and have,
\[\begin{align}
  & \Rightarrow 6{{x}^{2}}-30x+30=10{{x}^{2}}-60x+80 \\
 & \Rightarrow 0=10{{x}^{2}}-60x+80-6{{x}^{2}}+30x-30 \\
 & \Rightarrow 4{{x}^{2}}-30x+50=0 \\
\end{align}\]
Let us divide both side of the equation by 2 and have,
\[\Rightarrow 2{{x}^{2}}-15x+25=0\]
The above equation is in the general quadratic equation $a{{x}^{2}}+bx+c=0$ where $a=2,b=-15,c=25$. We use the splitting the middle term method in order to factorize it. The middle term in the equation $b=-15$ and we have to split in such a way the two split terms say $p,q$ when multiplied give the result $pq=a\times c=2\times 25=50.$ So the split terms will be the factors of 50. We have the prime factorization of 50 as
\[50=2\times 5\times 5\]
We choose the factors $2\times 5=10$ and 5 with negative sign so that their sum will be $p+q=-10+\left( -5 \right)=-15=b$. So we split the middle term as
\[\begin{align}
  & \Rightarrow 2{{x}^{2}}-10x-5x+25=0 \\
 & \Rightarrow 2x\left( x-5 \right)-5\left( x-5 \right)=0 \\
 & \Rightarrow \left( x-5 \right)\left( 2x-5 \right)=0 \\
\end{align}\]
We deduce from the above result that either $x-5=0$ and $2x-5=0$. Then we have either $x=5$ or $x=\dfrac{5}{2}$. So we have obtained two unequal and real roots of the given quadratic equation.\[\]


Note: We can alternatively convert the given equation in linear polynomial fractions by adding the polynomial fraction taking the least common multiple of denominators $\left( x-2 \right)$ and $\left( x-4 \right)$. We can alternatively factorize the quadratic equation $2{{x}^{2}}-15x+25=0$ using factor theorem where first we find a root by trial and error say $x=5$ and then divide the polynomial $2{{x}^{2}}-15x+25$ by $x-5$ to get the other factor.