Tag: theory of equations

Questions Related to theory of equations

Solve the equation: $x^{-2}-2x^{-1}=8$

reciprocal equations theory of equations maths

  1. $\dfrac{3}{4}, \dfrac{-1}{2}$

  2. $\dfrac{1}{4}, \dfrac{-1}{3}$

  3. $\dfrac{1}{3}, \dfrac{-1}{2}$

  4. $\dfrac{1}{4}, \dfrac{-1}{2}$

Show answer
Correct Option: D
Explanation:

Given, $x^{-2}-2x^{-1}=8$

$\Rightarrow \dfrac {1}{x^2}-\dfrac {2}{x}=8$
$\Rightarrow \dfrac {1-2x}{x^2}=8$
$\Rightarrow 1-2x=8x^2$
$\Rightarrow 8x^2+2x-1=0$
$\Rightarrow (2x+1)(4x-1)$
$\Rightarrow x=\dfrac {1}{4}, \dfrac {-1}{2}$

The number of solutions $(x, y, z)$ to the system of equations $ x + 2y + 4z = 9, 4yz + 2xz + xy = 13, xyz = 3 $ such that at least two of $ x, y, z$ are integers is

reciprocal equations theory of equations maths

  1. $3$

  2. $5$

  3. $6$

  4. $4$

Show answer
Correct Option: B
Explanation:
Let the roots of the system equation are:
$\alpha =x,\beta =2y,\gamma =4z$
$\alpha +\beta +\gamma =x+2y+4z=9$
$\alpha \beta +\beta \gamma +\gamma \alpha =2xy+8yz+yzx$
$=2(4yz+2xz+xy)\Rightarrow 26$
$\alpha \beta \gamma =8xyz\Rightarrow 24$
Thus,our polynomial should be:
$P^{3}-9P+26P-24=0$
$(P-2)(P-3)(P-4)=0$
since our roots are :
$\alpha =x,\beta =2y$ and $\gamma =4z$
$(x,2y,4z)=(2,3,4)$ or its permutations,or 6 combination.
However,note that one case if,
$x=4,2y=3$ and $4z=2$
$(x,y,z)=(4,\dfrac{3}{2},\dfrac{1}{2})$
which two of the roots are not an integer :Excluding of this case ,we have five solutions.

Solve the equation $\sqrt{4x^2-7x-15}-\sqrt{x^2-3x}=\sqrt{x^2-9}$

reciprocal equations theory of equations maths

  1. $2, 3$

  2. $1, 6$

  3. $-1, 3$

  4. $1, 3$

Show answer
Correct Option: D
Explanation:

Given equation is $\sqrt { 4{ x }^{ 2 }-7x-15 } =\sqrt { { x }^{ 2 }-9 } +\sqrt { { x }^{ 2 }-3x } $

$\Rightarrow \sqrt { x-3 } (\sqrt { 4x+5 } )=\sqrt { x-3 } (\sqrt { x+3 } +\sqrt { x } )$
Therefore $x=3$ is one solution and $\sqrt { 4x+5 } =\sqrt { x+3 } +\sqrt { x } $
By squaring above equation on both sides , we get $x+1=\sqrt{x(x+3)}$
Again square it on both sides , we get $x^{2}+2x+1=x^{2}+3x$
$\Rightarrow x=1$
Therefore option $D$ is correct

The roots of $a _ { 1 } x ^ { 2 } + b _ { 1 } x + c _ { 2 } = 0$ are reciprocal of the roots of the equation $a _ { 2 } x ^ { 2 } + b _ { 2 } x + c _ { 2 } = 0$

reciprocal equations theory of equations maths

  1. $\dfrac { a _ { 1 } } { a _ { 2 } } = \dfrac { b _ { 1 } } { b _ { 2 } } = \dfrac { c _ { 1 } } { c _ { 2 } }$

  2. $\dfrac { b _ { 1 } } { b _ { 2 } } = \dfrac { c _ { 1 } } { a _ { 2 } } = \dfrac { a _ { 1 } } { c _ { 2 } }$

  3. $\dfrac { a _ { 1 } } { a _ { 2 } } = \dfrac { b _ { 1 } } { c _ { 2 } } = \dfrac { c _ { 1 } } { b _ { 2 } }$

  4. $a _ { 1 } = \dfrac { 1 } { a _ { 2 } } , b _ { 1 } = \dfrac { 1 } { b _ { 2 } } , c _ { 1 } = \dfrac { 1 } { c _ { 2 } }$

Show answer
Correct Option: B
Explanation:
Given:${a} _{1}{x}^{2}+{b} _{1}x+{c} _{1}=0$     ........$(1)$
${a} _{2}{x}^{2}+{b} _{2}x+{c} _{2}=0$     ........$(2)$

Let $\alpha,\,\beta$ be the roots of ${a} _{1}{x}^{2}+{b} _{1}x+{c} _{1}=0$     ........$(1)$

$\Rightarrow\,\alpha+\beta=-\dfrac{{b} _{1}}{{a} _{1}}$ and 

$\alpha\beta=\dfrac{{c} _{1}}{{a} _{1}}$

Given:Roots of $(1)$ are reciprocal to $(2)$

$\dfrac{1}{\alpha}+\dfrac{1}{\beta}=-\dfrac{{b} _{2}}{{a} _{2}}$ and $\dfrac{1}{\alpha\beta}=\dfrac{{c} _{2}}{{a} _{2}}$

$\Rightarrow\,\dfrac{\alpha+\beta}{\alpha\beta}-\dfrac{{b} _{2}}{{a} _{2}}$ and $\dfrac{1}{\alpha\beta}=\dfrac{{c} _{2}}{{a} _{2}}$

Using $\alpha+\beta=-\dfrac{{b} _{1}}{{a} _{1}}$ and $\alpha\beta=\dfrac{{c} _{1}}{{a} _{1}}$ we have

$\Rightarrow\,\dfrac{-\dfrac{{b} _{1}}{{a} _{1}}}{\dfrac{{c} _{1}}{{a} _{1}}}=-\dfrac{{b} _{2}}{{a} _{2}}$ and $\dfrac{1}{\dfrac{{c} _{1}}{{a} _{1}}}=\dfrac{{c} _{2}}{{a} _{2}}$

$\Rightarrow\,\dfrac{-{b} _{1}}{{c} _{1}}=-\dfrac{{b} _{2}}{{a} _{2}}$ and
 
$\dfrac{{a} _{1}}{{c} _{1}}=\dfrac{{c} _{2}}{{a} _{2}}$

$\Rightarrow\,\dfrac{{b} _{1}}{{b} _{2}}=\dfrac{{c} _{1}}{{a} _{2}}$ and 

$\dfrac{{a} _{1}}{{c} _{1}}=\dfrac{{c} _{2}}{{a} _{2}}$

$\Rightarrow\,\dfrac{{b} _{1}}{{b} _{2}}=\dfrac{{c} _{1}}{{a} _{2}}$ and 

$\dfrac{{c} _{1}}{{a} _{1}}=\dfrac{{a} _{2}}{{c} _{2}}$

$\Rightarrow\,\dfrac{{b} _{1}}{{b} _{2}}=\dfrac{{c} _{1}}{{a} _{2}}$ and 

$\dfrac{{c} _{1}}{{a} _{2}}=\dfrac{{a} _{1}}{{c} _{2}}$

$\therefore\,\dfrac{{b} _{1}}{{b} _{2}}=\dfrac{{c} _{1}}{{a} _{2}}=\dfrac{{a} _{1}}{{c} _{2}}$

Option$(b)$ is correct.

Sum of roots is $-1$ and sum of their reciprocals is $\dfrac{1}{6}$, then equation is?

maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

  1. $x^2+x-6=0$

  2. $x^2-x+6=0$

  3. $6x^2+x+1=0$

  4. $x^2-6x+1=0$

Show answer
Correct Option: A
Explanation:

$\Rightarrow$  Let $\alpha$ and $\beta$ are roots of the equation.

According to the given condition,
$\Rightarrow$  $\alpha+\beta=-1$                             ------ ( 1 )
Again according to the given condition,
$\Rightarrow$  $\dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{1}{6}$

$\Rightarrow$  $\dfrac{\beta+\alpha}{\alpha\beta}=\dfrac{1}{6}$

$\Rightarrow$  $6(\alpha+\beta)=\alpha\beta$
$\Rightarrow$  $6(-1)=\alpha\beta$                          [ From ( 1 ) ]
$\therefore$  $\alpha\beta=-6$               ----  ( 2 )
Now, required equation,
$\Rightarrow$  $x^2-(\alpha+\beta)x+(\alpha\beta)=0$
Using ( 1 ) and ( 2 ) we get,
$\Rightarrow$  $x^2-(-1)x+(-6)=0$
$\therefore$  $x^2+x-6=0$

If the roots of $x^{3}-kx^{2}+14x-8=0$ are in geometric progression ,then $k=$

maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

  1. -3

  2. 7

  3. 4

  4. 0

Show answer
Correct Option: B
Explanation:

Let $\dfrac{a}{r}.a.ar $ be the roots 

$\Rightarrow \dfrac{a}{r}.a.ar=8$

$\Rightarrow a^{3}=8$

$\Rightarrow a=2$

$a=2 $ is a root given equation

$\Rightarrow 8-4k+28-8=0$

$\Rightarrow K=7$

The quadratic equation whose roots are twice the roots of  $2 x ^ { 2 } - 5 x + 2 = 0$  is:

maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

  1. $8 x ^ { 2 } - 10 x + 2 = 0$

  2. $x ^ { 2 } - 5 x + 4 = 0$

  3. $2 x ^ { 2 } - 5 x + 2 = 0$

  4. $x ^ { 2 } - 10 x + 6 = 0$

Show answer
Correct Option: B
Explanation:

$\begin{array}{l} Let\, \alpha \, and\, \beta \, be\, the\, root\, of\, the\, given\, equation. \ Then,\, \alpha +\beta =\frac { 5 }{ 2 } and\, \alpha \beta =\frac { 2 }{ 2 } =1 \ \therefore 2\alpha +2\beta  \ \therefore \left( { \alpha +\beta  } \right)  \ \therefore 5\left( { 2\alpha  } \right) \left( { 2\beta  } \right) =4 \ So\, the\, required\, equation\, is: \ { x^{ 2 } }-5x+4=0 \end{array}$


So, option $B$ is correct answer.

The sum and the product of the zeroes of a quadratic polynomial are $ \dfrac{-1}{2} $ and $ \dfrac{1}{2}$ respectively, then the polynomial is :

maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

  1. $2x^{2}+x+1$

  2. $2x^{2}-x+1$

  3. $2x^{2}-x-1$

  4. $2x^{2}+x-1$

Show answer
Correct Option: A
Explanation:

Given: Sum of zeroes $=-\dfrac 12$ and product of zeroes $=\dfrac 12$

We know,
$x^2-(\text{sum of zeroes})x+(\text{product of zeroes})=0$
$\Rightarrow x^2-\left(-\dfrac 12\right)x+\dfrac 12=0$
$\Rightarrow 2x^2+x+1=0$
is the required polynomial.

If $(b - c){x^2} + (c - a)x + (a - b) = 0$ has equal roots then $a,b,c$ are in :

maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

  1. A.P.

  2. G.P.

  3. H.P.

  4. none

Show answer
Correct Option: A
Explanation:
We have,

$(b-c)x^2+(c-a)x+(a-b)=0$

Comparing with the quadratic equation

$Ax^2+Bx+C=0$

$A=(b-c),B=c-a,C=a-b$

Discriminate when roots are equal

$D=B^2-4AC=0$

$D=(c-a)^2−4(b-c)(a-b)=0$

$D=(c^2+a^2−2ac)-4(ba-ac-b^2+bc)=0$

$D=c^2+a^2−2ac-4ab+4ac+4b^2-4bc=0$

$c^2+a^2+2ac-4b(a+c)+4b^2=0$

$(a+c)^2-4b(a+c)+4b^2=0$

$[(a+c)-2b]^2=0$

$a+c=2b$

So,

$a, b, c$ are in $A.P$

Hence, this is the answer.

If α+β=5α+β=5
State true or false.

maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

  1. True

  2. False

Show answer
Correct Option: A
Explanation:


Given
$\alpha^3 + \beta^3 = 35$
Sum of roots, $\alpha + \beta = 5$
Cubing both sides, 
$\alpha^3 + \beta^3 + 3\alpha\beta(\alpha + \beta) = 125$
$35 + 3\alpha\beta(5) = 125$
$15\alpha \beta =125-35$
$\alpha \beta =\frac { 90 }{ 15 } $
Product of roots, $\alpha\beta = 6$

The equation will be, $x^2 - Sx + P = 0$ will be
$x^2 - 5x + 6 = 0 $
Answer:1 as the given equation is true.