Tag: solving equations numerically

Questions Related to solving equations numerically

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

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. $170 { x }^{ 2 } { y }^{ 2 } { z }^{ 3 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\left(23{x}^{2}{y}^{2}z\right)\times\left(-15{x}^{3}y{z}^{2}\right)$
$=-345‬{x}^{2+3}{y}^{2+1}{z}^{1+2}$
$=-345{x}^{5}{y}^{3}{z}^{3}$
Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

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

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Rewrite as xy - 6x - 6y = 0, which factors to (x - 6)(y - 6) = 36. We need pairs (x-6, y-6) such that their product is 36 and x <= y. Possible pairs (u, v) where u*v=36 and u <= v: (-36, -1), (-18, -2), (-12, -3), (-9, -4), (-6, -6), (1, 36), (2, 18), (3, 12), (4, 9), (6, 6). There are 10 such pairs.

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

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

Reveal answer Fill a bubble to check yourself
D Correct answer
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, $8x\times 6x=8\times 20\times 6\times 20=19,200$

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

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

Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

Which of the following quadratics is irreducible?

  1. $2x^2 - 5x + 3$
  2. $2x^2 - 5x - 3$
  3. $5x^2 - 2x + 3$
  4. $5x^2 - 2x - 3$
Reveal answer Fill a bubble to check yourself
C Correct answer
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