Tag: finding roots by iteration

Questions Related to finding roots by iteration

x is ........... variable

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

  1. Dependent

  2. Independent

  3. None

  4. both

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Correct Option: B
Explanation:

Value of x does not depend on any other variable. so it is independent variable.

When multiplicity of a polynomial exist?

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

  1. when a factor appears in conjugate

  2. when a factor appears more than once

  3. when a factor appears more than twice

  4. None of the above

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Correct Option: B
Explanation:

"multiplicity of a polynomial exist when a factor of a polynomial repeats more than once"

For eg. In the polynomial $(x -2)^3(x-3)^2(x-1 )$ 
The root $2$ have multiplicity $3$, the root  $3$ have $2$, and the root $1$ have $1$
Hence, B is correct.

Let $R=gS-4$. When $S=8,R= 16$. When $S= 10$, R is

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

  1. 11

  2. 14

  3. 21

  4. 20

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Correct Option: C
Explanation:

$R=gS-4, S=8, R=16$
or $16=g\times 8-4$ or $16+4=g\times 8$
or $g=\frac {20}{8}=\frac {5}{2}$
Therefore, when $S=10$
$R=\frac {5}{2}\times 10-4=21$

A three-digit number beginning from the left is abc. The number is

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

  1. cba

  2. a+10b+100c

  3. 100a+10b+c

  4. none of these

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Correct Option: C
Explanation:

Here, c is the digit in the units' place; b standing in the tens' place represents b tens; similarly, a represents a hundreds.
The number is, therefore, equal to a hundreds + b tens + c units $= 100a + 10b + c$.

Condition for an irreducible quadratic equation is-

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

  1. discriminant is positive

  2. discriminant is negative

  3. discriminant is zero

  4. None of the above

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Correct Option: B
Explanation:

Irreducible quadratic equation can not be reduced more i.e the quadratic equation which do not have real roots   

This means the roots are imaginary
So if roots are imaginary, then discriminant $D  <  0$
Hence, option B is correct.

The factorized form of $x^5 +x^4-x-1$ is

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

  1. $(x+1)^2(x+1)^2(x-1)$

  2. $(x^2+1)(x^2+1)^2(x-1)$

  3. $(x^2+1)(x+1)^2(x-1)$

  4. $(x^2+1)^2(x-1)$

Show answer
Correct Option: C
Explanation:

$x^5 +x^4-x-1$
$=x^4(x + 1)-1(x + 1)$
$=(x +1)(x^4-1)$
$= (x + 1) (x^2+ 1) (x^2-1)$
$ = (x + 1) (x^2+ 1) (x + 1) (x -1)$
$ = (x^2+ 1) (x + 1)^2(x -1)$

Solve the simultaneous equations using the convergent iterations:
$5x$ + $y$ + $2z$ = $19$
$2x$ + $3y$ +$8z$ = $39$
$x$ + $4y$ -$2z$ = $-2$

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

  1. $x=2$, $y=2$ and $\text z =1$

  2. $x=2$, $y=1$ and $\text z =4$

  3. $x=3$, $y=2$ and $\text z =2$

  4. None of these

Show answer
Correct Option: B
Explanation:

$5x+y+2z=19$

$x+4y-2z=-2$

Adding above two equations we get,
$5x+y+2z+x+4y-2z=19-2$
$\implies 6x+5y=17$

Taking another equation 
$2x+3y+8z=39$

Multiplying the given last equation with $4$
$4x+16y-8z=-8$

Adding above two equations 
$2x+3y+8z+4x+16y-8z=31$
$\implies 6x+19y=31$

Now,
$6x+19y-(6x+5y)=31-17$
$\implies 19y-5y=14$
$\implies y=1$

By substituting the value of $y$ into the above equations 
we get,
$x=2$ and $z=4$
$\therefore\ x=2,y=1$ and $z=4$.

Solve the simultaneous equations using the convergent iterations:
$12x$ + $3y$ - $5z$ = $1$
$x$ + $5y$ +$3z$ = $28$
$3x$ + $7y$ $13z$ = $76$

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

  1. $x=1$, $y=4$ and $\text z =3$

  2. $x=1$, $y=3$ and $\text z =4$

  3. $x=2$, $y=3$ and $\text z =5$

  4. $x=3$, $y=2$ and $\text z =5$

Show answer
Correct Option: B

Find the answer to the equation $x^{3}$ - $2x$ = $25$  to one decimal place using trial and improvement method. 

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

  1. $3.7$

  2. $3.2$

  3. $3.6$

  4. $3.9$

Show answer
Correct Option: B
Explanation:

$f(x)={ x }^{ 3 }-2x=25$

$1.x=2$
${ x }^{ 3 }-2x=4$ (too small)

$2.\quad x=3\\ { x }^{ 3 }-2x=21$
$3.x=4{ x }^{ 3 }-2x=56$
$4.x<4\quad and\quad x>3$
$x=3.1$  
${ x }^{ 3 }-2x=23.6$ (close)
$ 5.x=3.2$
$ { x }^{ 3 }-2x=26.4$
$ x=3.2$

Use the Zero Product Property to solve the equation $(7x+2) (5x-4)=0$

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

  1. $\dfrac{2}{7}$ , $ \dfrac{-4}{5}$

  2. $\dfrac{-2}{7}$ , $ \dfrac{4}{5}$

  3. $\dfrac{-2}{5}$ , $ \dfrac{7}{5}$

  4. $\dfrac{2}{5}$ , $ \dfrac{-7}{5}$

Show answer
Correct Option: B
Explanation:

$(7x+2)(5x-4)=0\ (7x+2)=0\quad ;\quad (5x-4)=0\ x=-\dfrac { 2 }{ 7 } \quad ;\quad x=\dfrac { 4 }{ 5 } $