Question

Solve the following quadratic equation by factorization method: $6{{x}^{2}}-17ix-12=0$ \[\]

Answer 1

Hint: We put $-1={{i}^{2}}$ and convert the equation to $6{{x}^{2}}-17ix+12{{i}^{2}}=0$. We factorize with splitting the middle term by finding two numbers $p$ and $q$ such that $p+q=b=-17i$ and $pq=c\times a=12{{i}^{2}}\times 6=72{{i}^{2}}$. We can find $p,q$ taking combinations of factor from the prime factorization of 72.\[\]

Complete step by step answer:
We know that a quadratic equation with complex coefficient is always in the form $a{{x}^{2}}+bx+c=0\left( a\ne 0 \right)$ where $a,b,c$ complex numbers are. We solve a complex quadratic equation by factorization by finding two numbers $p,q$ such that $pq=\dfrac{c}{a}$ and $p+q=b$.
We are given the complex quadratic equation.
\[6{{x}^{2}}-17ix-12=0\]
We know that the complex number $i$ is the square root of negative unity that is $i=\sqrt{-1}$. So we have${{i}^{2}}=\sqrt{-1}\cdot \sqrt{-1}=\sqrt{\left( -1 \right)\left( -1 \right)}=-1$. Now we can write the given quadratic equation as follows;
\[\begin{align}
  & \Rightarrow 6{{x}^{2}}-17ix+12\left( -1 \right)=0 \\
 & \Rightarrow 6{{x}^{2}}-17ix+12{{i}^{2}}=0......\left( 1 \right) \\
\end{align}\]
We compare the above equation with the general quadratic equation $a{{x}^{2}}+bx+c=0$ and conclude $a=6,b=-17i,c=12{{i}^{2}}$.Now in order to factorize we need two numbers $p,q$ such that $p+q=b=-17i$and $pq=c\times a=12{{i}^{2}}\times 6=72{{i}^{2}}$. Let us find the prime factorization of 72 and try to find $p,q$. We have;
\[72=2\times 2\times 2\times 3\times 3\]
We see that if we can take $p=2\times 2\times 2=8$ and $q=3\times 3=9$ with negative signs multiplied by $i$ , they will satisfy our required conditions as follows;
\[\begin{align}
  & p+q=-8i+\left( -9i \right)=-17i=b \\
 & pq=\left( -8i \right)\left( -9i \right)=72{{i}^{2}}=c\times a \\
\end{align}\]
 So we can proceed from equation (1) as
\[\Rightarrow 6{{x}^{2}}-8xi-9xi+12{{i}^{2}}=0\]
We take $2x$ common form the first two terms and $-3i$ common from the second two terms and have;
\[\Rightarrow 2x\left( 3x-4i \right)-3i\left( 3x-4i \right)=0\]
We take $\left( 3x-4i \right)$ in the above step to have;
\[\Rightarrow \left( 3x-4i \right)\left( 2x-3i \right)=0\]
Since the product of two terms is zero then one of them has to be zero. So we have;
\[\begin{align}
  & \Rightarrow 3x-4i=0\text{ or }2x-3i=0 \\
 & \Rightarrow 3x=4i\text{ or }2x=3i \\
 & \Rightarrow x=\dfrac{4}{3}i\text{ or }x=\dfrac{3}{2}i \\
\end{align}\]

So the solutions of given complex quadratic equation are $x=\dfrac{4}{3}i,\dfrac{3}{2}i$.\[\]

Note: We note that when we are finding $p,q$ we have used trial and error with different combinations. A complex number is always in the form $a+ib$ where $a$ is the real part and $b$ is the complex part. If $a=0$ then the complex number is imaginary just like our obtained roots $x=\dfrac{4}{3}i,\dfrac{3}{2}i$. We note that the roots of a quadratic equation with complex coefficients are always complex while the roots of a quadratic equation with real coefficients may have real or complex roots.