
If \[ - 1 + i\] is a root of \[{x^4} + 4{x^3} + 5{x^2} + 2x + k = 0\] then the other roots are
A. \[ - 1\], \[ - 1\]
B. \[ - \dfrac{1}{2}\], \[ - \dfrac{3}{2}\]
C. \[ - 1 \pm \sqrt 2 \]
D. \[1 \pm \sqrt 2 \]
Answer
154.8k+ views
Hint: As we know that imaginary roots always occur in the form of conjugate pairs, we will find the conjugate pair and thus multiply both the roots and find the equation. Next, we will substitute \[x = - 1 + i\] in \[{x^4} + 4{x^3} + 5{x^2} + 2x + k = 0\] to determine the value of \[k\] and after getting the value of \[k\], we will factorize the equation and get the one root in the form of equation as \[{x^2} + 2x + 2\] and find the other factor also. Thus, we can find the remaining roots of the equation using \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Complete step by step solution
As given \[ - 1 + i\] is a root of the given equation, \[{x^4} + 4{x^3} + 5{x^2} + 2x + k = 0\]
We can determine the conjugate pair of \[ - 1 + i\].
That is \[ - 1 - i\].
Let \[x = - 1 + i\] and\[x = - 1 - i\], which can be written as \[x + 1 - i = 0\] and \[x + 1 + i = 0\] respectively.
Now, we will multiply both the equations obtained above and use \[{i^2} = - 1\] wherever needed.
Thus, we get,
\[
\Rightarrow \left( {x + 1 + i} \right)\left( {x + 1 - i} \right) = 0 \\
\Rightarrow x\left( {x + 1 - i} \right) + 1\left( {x + 1 - i} \right) + i\left( {x + 1 - i} \right) = 0 \\
\Rightarrow {x^2} + x - ix + x + 1 - i + ix + i - {i^2} = 0 \\
\Rightarrow {x^2} + 2x + 1 + 1 = 0 \\
\Rightarrow {x^2} + 2x + 2 = 0 \\
\]
Hence, the roots of the equation \[{x^2} + 2x + 2 = 0\] are \[ - 1 + i\] and \[ - 1 - i\].
Further, we will substitute \[x = - 1 + i\] in \[{x^4} + 4{x^3} + 5{x^2} + 2x + k = 0\] to evaluate the value of \[k\]
Use \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\] and \[{i^2} = - 1\] in the equation and solve for $k$.
Thus, we get,
\[
\Rightarrow {\left( {{{\left( { - 1} \right)}^2} + {i^2} - 2i} \right)^2} + 4\left( {{{\left( { - 1} \right)}^2} + {i^2} - 2i} \right)\left( { - 1 + i} \right) + 5\left( {{{\left( { - 1} \right)}^2} + {i^2} - 2i} \right) + 2\left( { - 1 + i} \right) + k = 0 \\
\Rightarrow {\left( {1 - 1 - 2i} \right)^2} + 4\left( {1 - 1 - 2i} \right)\left( { - 1 + i} \right) + 5\left( {1 - 1 - 2i} \right) + 2\left( { - 1 + i} \right) + k = 0 \\
\Rightarrow {\left( { - 2i} \right)^2} + 4\left( { - 2i} \right)\left( { - 1 + i} \right) + 5\left( { - 2i} \right) + 2\left( { - 1 + i} \right) + k = 0 \\
\Rightarrow - 4 - 8i\left( { - 1 + i} \right) - 10i - 2 + 2i + k = 0 \\
\Rightarrow - 4 + 8i - 8{i^2} - 10i - 2 + 2i + k = 0 \\
\Rightarrow - 4 + 8i + 8 - 10i - 2 + 2i + k = 0 \\
\Rightarrow 2 + k = 0 \\
\Rightarrow k = - 2 \\
\]
Thus, the given equation becomes,
\[{x^4} + 4{x^3} + 5{x^2} + 2x - 2 = 0\]
Next, factorize the above equation in the form such that one of the factors is \[{x^2} + 2x + 2\]:
\[
\Rightarrow {x^4} + 4{x^3} + 5{x^2} + 2x - 2 = 0 \\
\Rightarrow {x^2}\left( {{x^2} + 2x + 2} \right) + 2x\left( {{x^2} + 2x + 2} \right) - 1\left( {{x^2} + 2x + 2} \right) = 0 \\
\Rightarrow \left( {{x^2} + 2x - 1} \right)\left( {{x^2} + 2x + 2} \right) = 0 \\
\]
Apply zero-product rule on the obtained factors to solve for $x$.
\[ \Rightarrow {x^2} + 2x - 1 = 0\] or \[ \Rightarrow {x^2} + 2x + 2 = 0\]
Thus, for the factor \[{x^2} + 2x + 2 = 0\] we know the roots as \[x = - 1 + i\] and \[x = - 1 - i\].
So now we need to solve the equation \[{x^2} + 2x - 1 = 0\] to determine the roots by using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], where \[a = 1,{\text{ }}b = 2\] and \[c = - 1\].
Thus, we get,
\[
\Rightarrow x = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4 \cdot 1\left( { - 1} \right)} }}{{2\left( 1 \right)}} \\
\Rightarrow x = \dfrac{{ - 2 \pm \sqrt {4 + 4} }}{2} \\
\Rightarrow x = \dfrac{{ - 2 \pm \sqrt 8 }}{2} \\
\Rightarrow x = \dfrac{{ - 2 \pm 2\sqrt 2 }}{2} \\
\Rightarrow x = - \dfrac{2}{2} \pm \dfrac{{2\sqrt 2 }}{2} \\
\Rightarrow x = - 1 \pm \sqrt 2 \\
\]
Thus, the other roots of the given equation are \[x = - 1 \pm \sqrt 2 \]
Hence, the correct option is (3).
Note: As the given equation is of degree 4 so, we will have 4 roots of the equation. As one imaginary root is given and the conjugate of the given root is also the root of the equation, thus, we have 2 roots and need to find the other 2 roots of the equation. By substituting the value of \[x\] we can find the value of \[k\] and after that only we can determine the other 2 roots of the equation. As we can not find the factors using middle term splitting method so, we have used the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] to evaluate the other 2 roots.
Complete step by step solution
As given \[ - 1 + i\] is a root of the given equation, \[{x^4} + 4{x^3} + 5{x^2} + 2x + k = 0\]
We can determine the conjugate pair of \[ - 1 + i\].
That is \[ - 1 - i\].
Let \[x = - 1 + i\] and\[x = - 1 - i\], which can be written as \[x + 1 - i = 0\] and \[x + 1 + i = 0\] respectively.
Now, we will multiply both the equations obtained above and use \[{i^2} = - 1\] wherever needed.
Thus, we get,
\[
\Rightarrow \left( {x + 1 + i} \right)\left( {x + 1 - i} \right) = 0 \\
\Rightarrow x\left( {x + 1 - i} \right) + 1\left( {x + 1 - i} \right) + i\left( {x + 1 - i} \right) = 0 \\
\Rightarrow {x^2} + x - ix + x + 1 - i + ix + i - {i^2} = 0 \\
\Rightarrow {x^2} + 2x + 1 + 1 = 0 \\
\Rightarrow {x^2} + 2x + 2 = 0 \\
\]
Hence, the roots of the equation \[{x^2} + 2x + 2 = 0\] are \[ - 1 + i\] and \[ - 1 - i\].
Further, we will substitute \[x = - 1 + i\] in \[{x^4} + 4{x^3} + 5{x^2} + 2x + k = 0\] to evaluate the value of \[k\]
Use \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\] and \[{i^2} = - 1\] in the equation and solve for $k$.
Thus, we get,
\[
\Rightarrow {\left( {{{\left( { - 1} \right)}^2} + {i^2} - 2i} \right)^2} + 4\left( {{{\left( { - 1} \right)}^2} + {i^2} - 2i} \right)\left( { - 1 + i} \right) + 5\left( {{{\left( { - 1} \right)}^2} + {i^2} - 2i} \right) + 2\left( { - 1 + i} \right) + k = 0 \\
\Rightarrow {\left( {1 - 1 - 2i} \right)^2} + 4\left( {1 - 1 - 2i} \right)\left( { - 1 + i} \right) + 5\left( {1 - 1 - 2i} \right) + 2\left( { - 1 + i} \right) + k = 0 \\
\Rightarrow {\left( { - 2i} \right)^2} + 4\left( { - 2i} \right)\left( { - 1 + i} \right) + 5\left( { - 2i} \right) + 2\left( { - 1 + i} \right) + k = 0 \\
\Rightarrow - 4 - 8i\left( { - 1 + i} \right) - 10i - 2 + 2i + k = 0 \\
\Rightarrow - 4 + 8i - 8{i^2} - 10i - 2 + 2i + k = 0 \\
\Rightarrow - 4 + 8i + 8 - 10i - 2 + 2i + k = 0 \\
\Rightarrow 2 + k = 0 \\
\Rightarrow k = - 2 \\
\]
Thus, the given equation becomes,
\[{x^4} + 4{x^3} + 5{x^2} + 2x - 2 = 0\]
Next, factorize the above equation in the form such that one of the factors is \[{x^2} + 2x + 2\]:
\[
\Rightarrow {x^4} + 4{x^3} + 5{x^2} + 2x - 2 = 0 \\
\Rightarrow {x^2}\left( {{x^2} + 2x + 2} \right) + 2x\left( {{x^2} + 2x + 2} \right) - 1\left( {{x^2} + 2x + 2} \right) = 0 \\
\Rightarrow \left( {{x^2} + 2x - 1} \right)\left( {{x^2} + 2x + 2} \right) = 0 \\
\]
Apply zero-product rule on the obtained factors to solve for $x$.
\[ \Rightarrow {x^2} + 2x - 1 = 0\] or \[ \Rightarrow {x^2} + 2x + 2 = 0\]
Thus, for the factor \[{x^2} + 2x + 2 = 0\] we know the roots as \[x = - 1 + i\] and \[x = - 1 - i\].
So now we need to solve the equation \[{x^2} + 2x - 1 = 0\] to determine the roots by using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], where \[a = 1,{\text{ }}b = 2\] and \[c = - 1\].
Thus, we get,
\[
\Rightarrow x = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4 \cdot 1\left( { - 1} \right)} }}{{2\left( 1 \right)}} \\
\Rightarrow x = \dfrac{{ - 2 \pm \sqrt {4 + 4} }}{2} \\
\Rightarrow x = \dfrac{{ - 2 \pm \sqrt 8 }}{2} \\
\Rightarrow x = \dfrac{{ - 2 \pm 2\sqrt 2 }}{2} \\
\Rightarrow x = - \dfrac{2}{2} \pm \dfrac{{2\sqrt 2 }}{2} \\
\Rightarrow x = - 1 \pm \sqrt 2 \\
\]
Thus, the other roots of the given equation are \[x = - 1 \pm \sqrt 2 \]
Hence, the correct option is (3).
Note: As the given equation is of degree 4 so, we will have 4 roots of the equation. As one imaginary root is given and the conjugate of the given root is also the root of the equation, thus, we have 2 roots and need to find the other 2 roots of the equation. By substituting the value of \[x\] we can find the value of \[k\] and after that only we can determine the other 2 roots of the equation. As we can not find the factors using middle term splitting method so, we have used the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] to evaluate the other 2 roots.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
IIIT JEE Main Cutoff 2024

Photoelectric Effect and Stopping Potential with Work Function and Derivation for JEE

Newton’s Laws of Motion: Three Laws of Motion Explanation with Examples

Difference Between Metals and Non-Metals for JEE Main 2024

The stability of the following alkali metal chlorides class 11 chemistry JEE_Main

Lami's Theorem

Other Pages
JEE Advanced 2025 Revision Notes for Mechanics

Ideal and Non-Ideal Solutions Raoult's Law - JEE

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

List of Fastest Century In IPL - Cricket League and FAQs

NEET 2025: All Major Changes in Application Process, Pattern and More
