Question

Find the square root of \[{x^4} - 10{x^3} + 37{x^2} - 60x + 36\].
\[
  (A)\left| {\left( {{x^2} - 5x + 6} \right)} \right| \\
  (B)\left| {\left( {{x^2} + 5x + 6} \right)} \right| \\
  (C)\left| {\left( {{x^2} - 5x - 4} \right)} \right| \\
  (D)None{\text{ }}of{\text{ }}these \\
\]

Answer 1

Hint:
Bi- quadratic polynomials: Polynomials in which the highest degree is 4. It will be in the form of \[a{x^4} + b{x^3} + c{x^2} - dx + e\], where the leading coefficient (a) cannot be equal to 0. A bi-quadratic equation can have maximum 4 roots.

Complete step by step solution:
Given bi-quadratic equation: \[{x^4} - 10{x^3} + 37{x^2} - 60x + 36\]
As \[x = 2\] is satisfying the above bi-quadratic equation.
\[
   \Rightarrow {x^4} - 10{x^3} + 37{x^2} - 60x + 36 \\
   \Rightarrow {(2)^4} - 10{(2)^3} + 37{(2)^2} - 60(2) + 36 \\
   \Rightarrow 16 - 80 + 148 - 120 + 36 \\
   \Rightarrow 200 - 200 \\
\]
One of the factors of this bi-quadratic equation will be \[(x - 2)\].
On dividing the given bi-quadratic equation with \[(x - 2)\] we get;
\[ \Rightarrow \dfrac{{{x^4} - 10{x^3} + 37{x^2} - 60x + 36}}{{x - 2}} = (x - 2) \times ({x^3} - 8{x^2} + 21x - 18)\]……….EQ:01
Similarly, one of the solutions of the above cubic equation is again 2.
As \[x = 2\]is satisfying the above cubic equation.
\[
   \Rightarrow ({x^3} - 8{x^2} + 21x - 18) \\
   \Rightarrow {(2)^3} - 8{(2)^2} + 21(2) - 18 \\
   \Rightarrow 8 - 32 + 42 - 18 \\
   \Rightarrow 0 \\
\]
One of the factors of this cubic equation will be \[(x - 2)\].
On dividing the given cubic equation with \[(x - 2)\]we get;
\[ \Rightarrow \dfrac{{{x^3} - 8{x^2} + 21x - 18}}{{x - 2}} = (x - 2) \times ({x^2} - 6x + 9)\]…………EQ:02
Now, solving above quadratic equation using middle term splitting method, we get;
\[ \Rightarrow {x^2} - 6x + 9\]
\[
   \Rightarrow {x^2} - 3x - 3x + 9 \\
   \Rightarrow x(x - 3) - 3(x - 3) \\
   \Rightarrow (x - 3)(x - 3) \\
   \Rightarrow {(x - 3)^2} \\
\]
The above quadratic equation can be written as:
\[ \Rightarrow {x^2} - 6x + 9 = {(x - 3)^2}\]……… EQ:03
From EQ:01, EQ:02 and EQ:03, we get;
\[ \Rightarrow {x^4} - 10{x^3} + 37{x^2} - 60x + 36 = (x - 2)(x - 2){(x - 3)^2}\]
On square rooting both sides, we get;
\[
   \Rightarrow \sqrt {{x^4} - 10{x^3} + 37{x^2} - 60x + 36} = \sqrt {(x - 2)(x - 2){{(x - 3)}^2}} \\
   \Rightarrow \sqrt {{x^4} - 10{x^3} + 37{x^2} - 60x + 36} = \sqrt {{{(x - 2)}^2}{{(x - 3)}^2}} \\
   \Rightarrow \sqrt {{x^4} - 10{x^3} + 37{x^2} - 60x + 36} = \left| {(x - 2)(x - 3)} \right| \\
   \Rightarrow \sqrt {{x^4} - 10{x^3} + 37{x^2} - 60x + 36} = \left| {{x^2} - 3x - 2x + 6} \right| \\
   \Rightarrow \sqrt {{x^4} - 10{x^3} + 37{x^2} - 60x + 36} = \left| {{x^2} - 5x + 6} \right|
\]

Option (A) is correct.

Note:
Student’s find it very tricky to solve bi-quadratic equations or polynomials.
If one degree and three degree terms are not there in bi-quadratic equation, solve it with the help of variable change. Put \[{x^2} = t\] and then solve the transformed quadratic equation.
If all terms are present in the bi-quadratic equation, put \[1, - 1, 2, - 2, 3, - 3\]to satisfy the equation. In most cases, the equation will get satisfied by anyone of the above mentioned value.