Question

The number of real roots of the equation \[{x^4}\; + {\text{ }}\sqrt {{x^4}\; + {\text{ }}20} {\text{ }} = {\text{ }}22\]is
A. \[{\mathbf{4}}\]
B. $2$
C. $0$
D. $1$

Answer 1

Hint:A polynomial has that many numbers of roots as the highest power of the polynomial. The highest power of the polynomial is known as the degree of the polynomial. For example, \[3{x^2} - 5x + 7\]is the polynomial with the degree of the polynomial as \[2\] and thus, the total number of roots of the equation is also \[2\]. The roots of a particular equation in general can be real or non-real.

Formula Used:
Roots of a quadratic equation $ax^2+bx+c=0$ is
$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$


Complete step by step Solution:
We have given \[{x^4}\; + {\text{ }}\sqrt {{x^4}\; + {\text{ }}20} {\text{ }} = {\text{ }}22\]
Bring the square root term on the right-hand side of the equation and all the rest terms on the left-hand side, then square both sides.
\[{x^4}\; - 22{\text{ }} = - {\text{ }}\sqrt {{x^4}\; + {\text{ }}20} \]

Squaring both sides to remove the square root
\[{({x^4}\; - 22)^2}{\text{ }} = {x^4}\; + {\text{ }}20\]

Let \[{x^4} = y\]
\[{(y\; - 22)^2}{\text{ }} = y\; + {\text{ }}20\]
Solving the left-hand side by using algebraic identity
${y^2} + {22^2} - 2y(22) = y + 20$
\[{y^{2\;}} + {\text{ }}484{\text{ }}-{\text{ }}44y{\text{ }} = {\text{ }}y{\text{ }} + {\text{ }}20\]
\[{y^{2\;}} + {\text{ }}45y{\text{ }} + {\text{ }}464{\text{ }} = {\text{ }}0\]
\[{y^{2\;}}-{\text{ }}29y{\text{ }}-{\text{ }}16y{\text{ }} + {\text{ }}464{\text{ }} = {\text{ }}0\]
\[\;\left( {y{\text{ }}-{\text{ }}29} \right){\text{ }}\left( {y{\text{ }}-{\text{ }}16} \right){\text{ }} = {\text{ }}0\]
Put each bracket equal to zero \[\;\left( {y{\text{ }}-{\text{ }}29} \right) = {\text{ }}0\] and \[\;\left( {y{\text{ }}-{\text{ }}16} \right){\text{ }} = {\text{ }}0\]
Therefore, \[y{\text{ }} = {\text{ }}16\] and \[y{\text{ }} = 29\]
Put back y as \[{x^4}\]
\[{x^{4\;}} = {\text{ }}16\] and \[{x^4} = {\text{ }}29\]
Thus,\[\;x{\text{ }} = {\text{ }} \pm {\text{ }}2\] and \[x = {\text{ }} \pm {\text{ }}2.31\]
The number of roots obtained is 4 which are all real. Hence the number of real roots is equal to the maximum number of roots which is $2, - 2,2.31, - 2.31$

Hence, the correct option is (A).

Note: An equation with the highest power\[\;4\]is termed a quadratic equation whereas an equation with the highest power \[2\] is termed a quadratic equation. If the above question simply asked for the number of roots instead of real roots then we could answer the question without solving it.