Tag: theory of equations

Quizzes (6) Questions (104)

Questions Related to theory of equations

Multiple choice reciprocal equations theory of equations maths

The root of the reciprocal equation of second type and of odd degree is:

  1. $x=-1$
  2. $x=+1$
  3. $x=\pm1$
  4. $x=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$x = -1$ is a root of the reciprocal equation of first type and of odd degree. 
$x = 1$ is a root of the reciprocal equation of second type and of odd degree.
$x=±1$ are two roots of reciprocal equation of second type and of even degree. 
Multiple choice reciprocal equations theory of equations maths

lf $\mathrm{f}({x})=0$ is a reciprocal equation of second type and fifth degree, then a root of $\mathrm{f}({x})=0$  is:

  1. $0$
  2. $1$
  3. $-1$
  4. $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When the reciprocal equation is of an odd degree, second type, $x=1$ is always a solution.

Multiple choice reciprocal equations theory of equations maths

The roots equation $x^4-3x^3+4x^2-3x+1=0$ is 

  1. $0$
  2. $1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$x^4-3x^3+4x^2-3x+1=0$
This equation is resiprocal equation of first type as $a _{n-i}=a _i$
Dividing equation by $x^2$:
$x^2-3x+4-\dfrac3x+\dfrac{1}{x^2}=0$
$x^2+\dfrac{1}{x^2}-3x-\dfrac3x+4=0$
$\left(x+\dfrac1x\right)^2-2-3\left(x+\dfrac1x\right)+4=0$
$\left(x+\dfrac1x\right)^2-3\left(x+\dfrac1x\right)+2=0$
Let $x+\dfrac1x=y$
$y^2-3y+2=0$
$(y-1)(y-2)=0$
$y=1$ or $y=2$
For $y=1$:
 $x+\dfrac1x=1$
$x^2-x+1=0$
$D=(-1)^2-4(1)(1)=-3<0$
Hence no real value of x exists for this case.
For $y=2$:
$x+\dfrac1x=2$
$x^2-2x+1=0$
$(x-1)^2=0$
$x=1$
Hence solution of the given equation is x=1.

Multiple choice reciprocal equations theory of equations maths

If the coefficients from one end of an equation are equal in magnitude and sign to the coefficients from the other end, then the equation is said to be 

  1. reciprocal equation of second type

  2. reciprocal equation of first type

  3. reciprocal equation

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation


Multiple choice reciprocal equations theory of equations maths

Solve $(1-a^2)(x+a)-2a(1-x^2)=0$

  1. $x= a,\dfrac{-(1+a^2)}{2a}$
  2. $x= 0,\dfrac{-(1+a^2)}{2a}$
  3. $x= 0,\dfrac{-(1+a^2)}{4a}$
  4. $x= 2a,\dfrac{-(1+a^2)}{4a}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $(1-a^2)(x+a)-2a(1-x^2)=0$

$x+a-a^2x-a^3-2a+2ax^2=0$
$x-a-a^2x-a^3+2ax^2=0$
$2ax^2+x(1-a^2)-(a+a^3)=0$
By using formula, we have
$x=\dfrac {-(1-a^2)\pm \sqrt {(1-a^2)^2-4(2a)[-(a+a^3)]}}{2(2a)}$
$=\dfrac {-(1-a^2)\pm \sqrt {(1+3a^2)^2}}{4a}$
$=\dfrac {-(1-a^2)\pm (1+3a^2)}{4a}$
$=\dfrac {-1+a^2+1+3a^2}{4a}; \dfrac {-1+a^2-1-3a^2}{4a}$
$=\dfrac {4a^2}{4a}; \dfrac {-2-2a^2}{4a}$
$=a; \dfrac {-(1+a^2)}{2a}$

Multiple choice reciprocal equations theory of equations maths

Solve the reciprocal equation $x^4-3x^3+4x^2-3x+1=0$

  1. $0$
  2. $1$
  3. $3$
  4. $-1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We see that it is a reciprocal equation, so we divide it by $x^2$: 


$\Rightarrow$$x^2-3x+4-\dfrac{3}{x}+\dfrac{1}{x^2}=0$


$\Rightarrow$$(x^2+\dfrac{1}{x^2})-3(x+\dfrac{1}{x})+4=0$


We will now substitute $x+\dfrac{1}{x}=u$

By squaring it, and solve further we get

$\Rightarrow$$x^2+\dfrac{1}{x^2}=u^2-2$

We plug back into the equation to get

$\Rightarrow$$u^2-3u+2=0$

The solutions are $u _{1}=1, u _{2}=2$
So either $x+\dfrac{1}{x}=1$ or $2$

From the first solution we get 
$\Rightarrow$$x^2-x+1=0$    which has no solution
From the second one we get
$\Rightarrow$$x^2-2x+1=0$ 
$\Rightarrow$$x=1$

Multiple choice reciprocal equations theory of equations maths

If the coefficients from one end of an equation are equal in magnitude and opposite in sign to the coefficients from the other end, then the equation is said to be 

  1. reciprocal equation of second type

  2. reciprocal equation of first type

  3. reciprocal equation

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Consider a general equation:

$a _{n}x^n+a _{n-1}x^{n-1}+a _{n-2}x^{n-2}+....................+a _1x+a _0=0$

Now if  $a _n-i=-a _i ,$        $  i=0,1,2,3,...........n$

Then this type of equation is called as reciprocal equation of second type.

Ex- $4x^4-9x^3+9x-4=0$
Multiple choice reciprocal equations theory of equations maths

An equation of the form $2x^4-3x^3+7x^2-3x+2=0$ is called a .................

  1. Reciprocal equation

  2. Radical equation

  3. Exponential equation

  4. Quadratic equation
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As 'The coefficients from beginning to end and vice versa are the same'
therefore given equation is a reciprocal equation.

Multiple choice reciprocal equations theory of equations maths

Solve for $x$:   $\dfrac{8\sqrt{x-5}}{3x-7}=\dfrac{\sqrt{3x-7}}{x-5}$

  1. $13$
  2. $23$
  3. $14$
  4. $-13$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\cfrac{8\sqrt{x-5}}{3x-7}=\cfrac{\sqrt{3x-7}}{x-5}$

$\Rightarrow 8(x-5)^{\tfrac 32}=(3x-7)^{\tfrac 32}$
Squaring on both sides we get,
$\Rightarrow 64(x-5)^{3}=(3x-7)^{3}$
Taking cube roots on both sides, we get
$\Rightarrow 4(x-5)=(3x-7)$
$\Rightarrow 4x-20=3x-7$
$\Rightarrow x=13$
Hence, option A is correct.

Multiple choice reciprocal equations theory of equations maths

Find $x$,  $2^{x^2}:2^{2x}=8:1$

  1. $3,-1$
  2. $3,1$
  3. $-3,-1$
  4. $-3,1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $2^{x^2}:2^{2x}=8:1$

$\Rightarrow \dfrac {2{x^2}}{2^{2x}}=\dfrac {8}{1}$
$\Rightarrow 2^{x^2}=8.2^{2x}$
$\Rightarrow 2^{x^2}=2^3.2^{2x}$
$\Rightarrow 2^{x^2}=2^{2x+3}$
$\Rightarrow x^2=2x+3$ ....As bases are equal, powers must be equal
$\Rightarrow x^2-2x-3=0$
$\Rightarrow (x-3)(x+1)=0$
$\therefore x=3,-1$