Tag: finding roots by iteration

Questions Related to finding roots by iteration

The product of $\left( { 23 x }^{ 2 }{ y }^{ 2 }z \right)$ and $\left( -15{ x }^{ 3 }{ yz }^{ 2 } \right) $ is .........................  .

  1. $-345 { x }^{ 5 } { y }^{ 3 } { z }^{ 3 }$

  2. $345 { x }^{ 2 } { y }^{ 2 } { z }^{ 3 }$

  3. $145 { x }^{ 2 } { y }^{ 2 } { z }^{ 3 }$

  4. $170 { x }^{ 2 } { y }^{ 2 } { z }^{ 3 }$


Correct Option: A
Explanation:
$\left(23{x}^{2}{y}^{2}z\right)\times\left(-15{x}^{3}y{z}^{2}\right)$
$=-345‬{x}^{2+3}{y}^{2+1}{z}^{1+2}$
$=-345{x}^{5}{y}^{3}{z}^{3}$

The number of integers (positive, negative or zero) solutions of
$xy-6(x+y)=0$ with x is less than or equal to y is:

  1. 5

  2. 10

  3. 12

  4. none of these


Correct Option: A

The degree of polynomial $p(x)=x^ {2}-3x-4x^ {3}-6$ is

  1. $2$

  2. $1$

  3. $3$

  4. $6$


Correct Option: C
Explanation:

Given : $p{\left( x \right)} = {x}^{2} - 3x - 4 {x}^{3} - 6$


Since the greatest exponent in $p{\left( x \right)}$ is $3$, thus the degree of $p{\left( x \right)}$ is $3$.

The method of finding solution by trying out various values for the variable is called

  1. Rrror method

  2. Trial and error method

  3. Testing method

  4. Checking method


Correct Option: B
Explanation:

The required method is called "Trial and error method"

If $a\times b=\frac {a}{b}+\frac {b}{a}-ab$, then the value of $1^*2$ is

  1. 2

  2. 1/2

  3. 2/3

  4. 1


Correct Option: B
Explanation:

$1^*2=\frac {1}{2}+\frac {2}{1}-1\times 2=\frac {1+4-4}{2}=\frac {1}{2}$

If $(8x)^2 + (6x)^2 = d^2$ and $d = 200$, then $8x \times 6x$ is equal to

  1. 18,200

  2. 18,500

  3. 18,900

  4. 19,200


Correct Option: D
Explanation:

$(8x)^2+(6x)^2=d^2$
or $64x^2+36x^2=d^2$
or $100x^2=d^2$
or $d=\sqrt {100x^2}=10x$
$10x=200$
$\therefore x=\frac {200}{10}=20$
Hence, $8x\times 6x=8\times 20\times 6\times 20=19,200$

If $2\pi rh = 2\pi r^2$ and $h = 5$, then r is equal to

  1. 5

  2. 110

  3. 15

  4. 12


Correct Option: A
Explanation:

$2\pi r h=2\pi r^2$
$2\pi \times r\times 5=2\pi r^2$ or $r=5$

Is the following quadratic polynomial reducible or irreducible?
$f(x) = -2x^2-2x-1$

  1. Reducible with one real root

  2. Reducible with two real roots

  3. Irreducible

  4. None of these


Correct Option: C
Explanation:

To check whether the given quadratic polynomial is reducible or irreducible, we need to calculate the discriminant

Calculate the discriminant for the equation, $-2x^2-2x-1=0$

$D=b^{2}-4ac=(-2)^2-4(-2)(-1)=-4<0$
Quadratic equation is irreducible if $D<0$
$\therefore$ The quadratic polynomial is irreducible.

Correct option is C

If $\dfrac {1}{x}-\dfrac {1}{y}=\dfrac {1}{z}$, then z is equal to

  1. y-x

  2. x-y

  3. $\dfrac {y-x}{xy}$

  4. $\dfrac {xy}{y-x}$


Correct Option: D
Explanation:

$\dfrac {1}{x}-\dfrac {1}{y}=\dfrac {1}{z}$ or $\dfrac {y-x}{xy}=\dfrac {1}{z}$ or $\dfrac {z}{1}=\dfrac {xy}{y-x}$

Which of the following quadratics is irreducible?

  1. $2x^2 - 5x + 3$

  2. $2x^2 - 5x - 3$

  3. $5x^2 - 2x + 3$

  4. $5x^2 - 2x - 3$


Correct Option: C
Explanation:

We can check by comparing the D of quadratic equations with their relation with 0


$2x^2-5x+3=0$
$D=(-5)^2-4(2)(3)=25-24=1>0$
Reducible

$2x^2-5x-3=0$
$D=(-5)^2-4(2)(-3)=25+24=49>0$
Reducible

$5x^2-2x+3=0$
$D=(-2)^2-4(5)(3)=4-60=-56<0$
Irreducible


$5x^2-2x-3=0$
$D=(-2)^2-4(5)(-3)=4+60=64>0$
Rreducible


Therefore correct option is C