Tag: finding roots by iteration

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

When multiplicity of a polynomial exist?

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. cba

  2. a+10b+100c

  3. 100a+10b+c

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $= 100a + 10b + c$.

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

Condition for an irreducible quadratic equation is-

  1. discriminant is positive

  2. discriminant is negative

  3. discriminant is zero

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $z=4$.

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

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

  1. $3.7$
  2. $3.2$
  3. $3.6$
  4. $3.9$
Reveal answer Fill a bubble to check yourself
B Correct answer
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$

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

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

  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}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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