Tag: reciprocal equations

Questions Related to reciprocal equations

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

reciprocal equations theory of equations maths

  1. $x=-1$

  2. $x=+1$

  3. $x=\pm1$

  4. $x=0$

Show answer
Correct Option: B
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. 

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

reciprocal equations theory of equations maths

  1. $0$

  2. $1$

  3. $-1$

  4. $2$

Show answer
Correct Option: B
Explanation:

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

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

reciprocal equations theory of equations maths

  1. $0$

  2. $1$

  3. $2$

  4. $3$

Show answer
Correct Option: B
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.

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 

reciprocal equations theory of equations maths

  1. reciprocal equation of second type

  2. reciprocal equation of first type

  3. reciprocal equation

  4. None of these

Show answer
Correct Option: B
Explanation:


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

reciprocal equations theory of equations maths

  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. $x= 2a,\dfrac{-(1+a^2)}{4a}$

Show answer
Correct Option: A
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
$x=\dfrac {-(1-a^2)\pm \sqrt {(1-a^2)^2-4(2a)[-(a+a^3)]}}{2(2a)}$
$=\dfrac {-(1-a^2)\pm \sqrt {(1+3a^2)^2}}{4a}$
$=\dfrac {-(1-a^2)\pm (1+3a^2)}{4a}$
$=\dfrac {-1+a^2+1+3a^2}{4a}; \dfrac {-1+a^2-1-3a^2}{4a}$
$=\dfrac {4a^2}{4a}; \dfrac {-2-2a^2}{4a}$
$=a; \dfrac {-(1+a^2)}{2a}$

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

reciprocal equations theory of equations maths

  1. $0$

  2. $1$

  3. $3$

  4. $-1$

Show answer
Correct Option: B
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$

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 

reciprocal equations theory of equations maths

  1. reciprocal equation of second type

  2. reciprocal equation of first type

  3. reciprocal equation

  4. None of these

Show answer
Correct Option: A
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- $4x^4-9x^3+9x-4=0$

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

reciprocal equations theory of equations maths

  1. Reciprocal equation

  2. Radical equation

  3. Exponential equation

  4. Quadratic equation

Show answer
Correct Option: A
Explanation:

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

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

reciprocal equations theory of equations maths

  1. $13$

  2. $23$

  3. $14$

  4. $-13$

Show answer
Correct Option: A
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.

Find $x$,  $2^{x^2}:2^{2x}=8:1$

reciprocal equations theory of equations maths

  1. $3,-1$

  2. $3,1$

  3. $-3,-1$

  4. $-3,1$

Show answer
Correct Option: A
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
$\Rightarrow x^2-2x-3=0$
$\Rightarrow (x-3)(x+1)=0$
$\therefore x=3,-1$