Tag: fundamental theorem of algebra

Questions Related to fundamental theorem of algebra

The number of ordered pairs of integers (x,y) satisfying the equation
${ x }^{ 2 }+6x+{ y }^{ 2 }=4$ is

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. 2

  2. 4

  3. 6

  4. 8

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Correct Option: D
Explanation:
Given,

$x^2+6x+y^2=4$ 

$\Rightarrow x^2+6x+9+y^2=4+9$

$\Rightarrow x^2+3^2+2(3x)+y^=13$

$\Rightarrow (x+3)^2+y^2=13$

sum of two squares is $13$

$\therefore $ when $(x+3)^2=9,x=0,-6$ and $y^2=2,-2$ $\Rightarrow 4$ ordered pairs.

$\therefore $ when $(x+3)^2=4,x=-1,-5$ and $y^2=3,-3$ $\Rightarrow 4$ ordered pairs.

A total of $8$ pairs.

The solution set of the system of equations $\log _{ 3 }{ x } +\log _{ 3 }{ y } =2+\log _{ 3 }{ 2 } \quad and\quad \log _{ 27 }{ (x+y) } =\dfrac { 2 }{ 3 } $ is :

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. {6,3}

  2. {3,6}

  3. {6,12}

  4. {12,6}

Show answer
Correct Option: A
Explanation:
Given,

$\log _3\left(x\right)+\log _3\left(y\right)=2+\log _3\left(2\right)$

$\log _3\left(x\right)=2+\log _3\left(2\right)-\log _3\left(y\right)$

$x=3^{2+\log _3\left(2\right)-\log _3\left(y\right)}$

$=3^{\log _3\left(2\right)}\cdot \:3^{-\log _3\left(y\right)}\cdot \:3^2$

$=2\cdot \:3^{-\log _3\left(y\right)}\cdot \:3^2$

$=2y^{-1}\cdot \:3^2$

$\Rightarrow x=\dfrac{18}{y}$.......(1)

Now,

$\log _{27}\left(x+y\right)=\dfrac{2}{3}$

from (1)

$\log _{27}\left(\dfrac{18}{y}+y\right)=\dfrac{2}{3}$

$\dfrac{18}{y}+y=27^{\frac{2}{3}}$

$\Rightarrow 18+y^2=9y$

$y^2-9y+18=0$

$(y-3)(y-6)=0$

$\therefore y=3,6$

$x=\dfrac{18}{y}=\dfrac{18}{3}=6$

$x=\dfrac{18}{y}=\dfrac{18}{6}=3$

$(x,y)=\left \{ 6,3 \right \}or\left \{ 3,6 \right \}$

Number of real solutions of the equation $\sqrt { \log _{ 10 }{ (-x) }  } =\log _{ 10 }{ \sqrt { { x }^{ 2 } }  } $ is :

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. zero

  2. exactly 1

  3. exactly 2

  4. 5

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Correct Option: A
Explanation:
Given,

$\sqrt{\log _{10}(-x)}=\log _{10}(\sqrt{x^2})$

$\sqrt{\log _{10}(-x)}=\log _{10}x$

$\log _{10}\left(-x\right)=\left(\log _{10}\left(x\right)\right)^2$

$\log _{10}\left(-1\right)+\log _{10}\left(x\right)=\left(\log _{10}\left(x\right)\right)^2$

let $\log _{10}\left(x\right)=u$

$\log _{10}\left(-1\right)+u=\left(u\right)^2........\log _{10}(-1) \ is \ not \ defined$

$u=\mathrm{Undefined}$

$\:\log _{10}\left(x\right)=\mathrm{Undefined}:\quad x=10^{\mathrm{Undefined}}$

$x=10^{\mathrm{Undefined}}\space\mathrm{False}$

No solution for $\:x\in \mathbb{R}$

Number of solutions satisfying, $\sqrt { 5-{ log } _{ 2 }x } =3-{ log } _{ 2 }x$ are :

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. 1

  2. 2

  3. 3

  4. 4

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Correct Option: A
Explanation:

$\sqrt { 5-{ \log } _{ 2 }x } =3-{ \log } _{ 2 }x$                  ---- ( 1 )


Let $\log _2x=y$                     ----- ( 2 )


$\Rightarrow$  $\sqrt{5-y}=3-y$

Squaring both sides, 

$\Rightarrow$  $5-y=9-6y+y^2$

$\Rightarrow$  $y^2-5y+4=0$

$\Rightarrow$  $y^2-4y-y+4=0$

$\Rightarrow$  $y(y-4)-1(y-4)=0$

$\Rightarrow$  $(y-4)(y-1)=0$

$\Rightarrow$  $y=4$ and $y=1$

Substituting $y=1$ in ( 1 ) we get,

$\Rightarrow$  $\log _2x=1$

We know, $\log _ba=x\Rightarrow a=b^x$

$\therefore$  $x=2^1=2$

Substituting $y=4$ in ( 1 ) we get,

$\Rightarrow$  $\log _2x=4$

We know, $\log _ba=x\Rightarrow a=b^x$

$\therefore$  $x=2^4=16$

Substituting $\log _2x=1$ in ( 1 ) we get,

$\sqrt{5-1}=3-1$

$\Rightarrow$  $\sqrt{4}=2$

$\therefore$  $2=2$

Substituting $\log _2x=4$ in ( 1 ) we get,

$\sqrt{5-4}=3-4$

$\Rightarrow$  $\sqrt{1}=-1$

$\therefore$  $1=-1$

Hence, we can see only $\log _2x=1$ satisfying.

$\therefore$  $\sqrt { 5-{ \log } _{ 2 }x } =3-{ \log } _{ 2 }x$ has only $1$ solution.

Number of ordered pair(s) of (x,y) satisfying the system of equations, $\log _2 xy = 5$ and $\log _{\frac{1}{2}} \frac{x}{y} = 1$ is:

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. one

  2. two

  3. three

  4. four

Show answer
Correct Option: B
Explanation:
Given,

$\log _2\left(xy\right)=5,\:\log _{\frac{1}{2}}\left(\frac{x}{y}\right)=1$

$\log _2\left(xy\right)=5$

$xy=2^5=32$

$x=\dfrac{32}{y}$

$\log _{\frac{1}{2}}\left(\dfrac{x}{y}\right)=1$

$\log _{\frac{1}{2}}\left(\dfrac{\frac{32}{y}}{y}\right)=1$

$\dfrac{\frac{32}{y}}{y}=\left(\dfrac{1}{2}\right)^1$

$\dfrac{32}{y}=\dfrac{1}{2}y$

$y^2=64$

$\Rightarrow y=\pm 8$

$x=\dfrac{32}{\pm 8}=\pm 4$

$(4,8),(-4,-8)$ Therefore $2$ ordered pairs

If $H.C.F. \left(a,b\right) = 9$ and $a  . b = 100$, then $L.C.M.\left(a,b\right) =$

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. $200$

  2. $100/3$

  3. $100/9$

    1. $150$
Show answer
Correct Option: C
Explanation:

$H.C.F.(a,b)=9\ a.b=100\ L.C.M.(a,b)=\frac { a.b }{ H.C.F } \ =\quad \frac { 100 }{ 9 } $

Find multiplicity of the polynomial
$f(x) = (x-1)^2(2x+5)^3(x^2+1)^2(x+\pi^2)^4$

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. $4$

  2. $2$

  3. $8$

  4. $1$

Show answer
Correct Option: A
Explanation:

The roots of the given function are $1,-\frac{5}{2} ,-i,i,-\pi^{2}$

The multiplicity of$x=1$ is $2$
The multiplicity of $x=-\frac{5}{2}$ is $3$
The multiplicity of $x=i $ is $2$
The multiplicity of $x=-i$ is $2$
The multiplicity of $x=-\pi^{2}$ is $4$
Therefore the multiplicity of polynomial is $4$
Therefor option $A$ is correct

The multiplicity of the root $x=1$ for the function $f(x) = x^2(x+1)^3(x-2)^2(x-1)$ is

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. $1$

  2. $2$

  3. $3$

  4. $4$

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Correct Option: A
Explanation:

After factorising $f(x)$, $(x-1)$ is appearing only once 

Multiplicity of the root is the number of times a number is zero of a polynomial.
Here $x=1$ is the zero of $f(x)$ only once as it is appearing one.
So, the multiplicity of the root $x=1$x=1 for the function is one.
Hence, option A is correct.

List the multiplicities of the zeroes of the polynomial $P(x)=x^2-14x+49$

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. $x=5$ is a zero of multiplicity $3$.

  2. $x=7$ is a zero of multiplicity $2$.

  3. $x=6$ is a zero of multiplicity $1$.

  4. None of these

Show answer
Correct Option: B
Explanation:

Given, $x^2-14x+49$

$\Rightarrow x^2-7x-7x+49$
$\Rightarrow x(x-7)-7(x-7)$
$\Rightarrow (x-7)^2$
So, $x=7$ is a zero of multiplicity $2$.

Is the following quadratic polynomial reducible or irreducible?
$f(x) = x^2 - \sqrt2$

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. Reducible with one real root

  2. Reducible with two real roots

  3. Irreducible

  4. None of these

Show answer
Correct Option: B
Explanation:

Let $\sqrt 2=a^2\Rightarrow a^2=2^{1/2}\Rightarrow a=2^{1/4}$

So $f(x)=x^2-a^2=(x+a)(x-a)=(x+2^{1/4})(x-2^{1/4})$, using $a^2-b^2=(a+b)(a-b)$
So given quadratic is reducible to two real roots