Question

The following equation \[12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=0\] has
(a). four real and distinct roots
(b). two irrational roots
(c). one integer root
(d). two imaginary roots

Answer 1

Hint: Divide the whole expression by \[{{x}^{2}}\]. Simplify and rearrange the expression. Put \[\left( x+\dfrac{1}{x} \right)\] as y. Thus, solve the quadratic equations formed in y. Substitute and solve the 2 quadratic equations formed by x and get the roots of x.

Complete step-by-step answer:
Given us the equation, \[12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12=0-(1)\]
Now let us divide the entire equation by \[{{x}^{2}}\].
\[\dfrac{12{{x}^{4}}-56{{x}^{3}}+89{{x}^{2}}-56x+12}{{{x}^{2}}}=0\]
Thus we get the equation as,
\[12{{x}^{2}}-56x+89-\dfrac{56}{x}+\dfrac{12}{{{x}^{2}}}=0-(2)\]
Now let us rearrange the above equation,
\[\begin{align}
  & 12{{x}^{2}}+\dfrac{12}{{{x}^{2}}}-56x-\dfrac{56}{x}+89=0 \\
 & 12\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)-56\left( x+\dfrac{1}{x} \right)+89=0 \\
\end{align}\]
Let us add and subtract 2 in the first term,
\[12\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}}+2-2 \right)-56\left( x+\dfrac{1}{x} \right)+89=0\]
We know, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\].
\[\begin{align}
  & \therefore 12\left[ \left( {{x}^{2}}+2+\dfrac{1}{{{x}^{2}}} \right)-2 \right]-56\left[ x+\dfrac{1}{x} \right]+89=0 \\
 & 12\left[ {{\left( x+\dfrac{1}{x} \right)}^{2}}-2 \right]-56\left( x+\dfrac{1}{x} \right)+89=0 \\
 & 12{{\left( x+\dfrac{1}{x} \right)}^{2}}-\left( 12\times 2 \right)-56\left( x+\dfrac{1}{x} \right)+89=0 \\
 & 12{{\left( x+\dfrac{1}{x} \right)}^{2}}-56\left( x+\dfrac{1}{x} \right)+89-24=0 \\
 & \therefore 12{{\left( x+\dfrac{1}{x} \right)}^{2}}-56\left( x+\dfrac{1}{x} \right)+65=0-(3) \\
\end{align}\]
Now let us put, \[x+\dfrac{1}{x}=y\] in equation (3).
Thus we get a simplified expression as,
\[12{{y}^{2}}-56y+65=0\]
This expression is similar to the quadratic equation, \[a{{y}^{2}}-by+c=0\].
Thus comparing both the equation, we get
a = 12, b = -56, c =65
Substitute these values in the quadratic formula,
\[y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-\left( -56 \right)\pm \sqrt{{{\left( -56 \right)}^{2}}-4\times 12\times 65}}{2\times 12}\]
\[\begin{align}
  & y=\dfrac{56\pm \sqrt{3136-3120}}{24}=\dfrac{56\pm \sqrt{16}}{24} \\
 & y=\dfrac{56\pm 4}{24} \\
\end{align}\]
\[\therefore y=\dfrac{56+4}{24}\] and \[y=\dfrac{56-4}{24}\]
\[y=\dfrac{60}{24}=\dfrac{5}{2}\] and \[y=\dfrac{52}{24}=\dfrac{13}{6}\].
Thus we got \[y=\dfrac{5}{2}\] and \[y=\dfrac{13}{6}\].
Thus, \[x+\dfrac{1}{x}=y\] ,substitute the value of y.
\[x+\dfrac{1}{x}=\dfrac{5}{2}\] and \[x+\dfrac{1}{x}=\dfrac{13}{6}\].
Cross multiply and solve the quadratic equation.
\[\dfrac{{{x}^{2}}+1}{2}=\dfrac{5}{2}\] and \[\dfrac{{{x}^{2}}+1}{2}=\dfrac{13}{6}\].
\[2\left( {{x}^{2}}+1 \right)=5x\] and \[6\left( {{x}^{2}}+1 \right)=13x\]
\[2{{x}^{2}}-5x+2=0\] and \[6{{x}^{2}}-13x+6=0\].
Both the equations are similar to the quadratic equation, \[a{{x}^{2}}+bx+c=0\].
Thus for \[2{{x}^{2}}-5x+2=0\]
a = 2, b = -5, c = 2
And for \[6{{x}^{2}}-13x+6=0\]
a = 6 , b = -13, c = 6
Substituting in quadratic formula, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
\[\begin{align}
  & x=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\times 2\times 2}}{2\times 2} \\
 & x=\dfrac{5\pm \sqrt{25-16}}{4} \\
 & x=\dfrac{5\pm \sqrt{9}}{4}=\dfrac{5\pm 3}{4} \\
\end{align}\]
Thus, we got \[x=\dfrac{5+3}{4}=2\]
\[x=\dfrac{5-3}{4}=\dfrac{2}{4}=\dfrac{1}{2}\]
\[\begin{align}
  & x=\dfrac{-\left( -13 \right)\pm \sqrt{{{\left( -13 \right)}^{2}}-4\times 6\times 6}}{2\times 6} \\
 & x=\dfrac{13\pm \sqrt{169-144}}{12} \\
 & x=\dfrac{13\pm \sqrt{25}}{12}=\dfrac{13\pm 5}{12} \\
\end{align}\]
Here, \[x=\dfrac{13+5}{12}=\dfrac{18}{12}=\dfrac{3}{2}\]
\[x=\dfrac{13-5}{12}=\dfrac{8}{12}=\dfrac{2}{3}\]
Thus we got four values of \[x=2,\dfrac{1}{2},\dfrac{3}{2},\dfrac{2}{3}\].
\[\therefore \] The equation has 4 real and distinct roots and 2 is an integer root.
\[\therefore \] Option (a) and (c) is the correct answer.

Note: Remember to divide the entire expression by \[{{x}^{2}}\], so as to simplify the expression. Irrational roots are the roots that cannot be expressed in \[\dfrac{p}{q}\] form. Here we have three rational roots that can be expressed in \[\dfrac{p}{q}\] form i.e. \[\dfrac{1}{2},\dfrac{3}{2}\] and \[\dfrac{2}{3}\].