Tag: theory of equations

Questions Related to theory of equations

If $P ( \alpha , \beta )$ moves on $x ^ { 2 } + y ^ { 2 } - 2 x + 6 y + 1 = 0$ then minimum value of $a ^ { 2 } + \beta ^ { 2 } - 2 a - 4 \beta$ is 

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

  1. -3

  2. -1

  3. 1

  4. 3

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Correct Option: B

The sum and the product of zeroes of a quadratic polynomial $p(x)$ are $-7$ and $-10$ respectively. Then $p(x)$ is :

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

  1. $x^{2}-7x-10$

  2. $x^{2}-7x+10$

  3. $x^{2}+7x-10$

  4. $x^{2}+7x+10$

Show answer
Correct Option: C
Explanation:
Given: Sum of zeroes $=-7$ and product of zeroes $=-10$
We know that
$p(x)=x^2-(\text{sum of zeroes})x+(\text{product of zeroes})$
$\Rightarrow p(x)=x^2-(-7)x+(-10)$
$\Rightarrow p(x)=x^2+7x-10$
is the required polynomial.

If $\alpha$ and $\beta$ are the roots of the equation $ax^{2} \, + \, bx \, + \, c \, = \, 0$. The equation whose roots are as given below.
$\alpha \, + \,\dfrac{1}{\beta} \, , \, \beta \, + \, \dfrac{1}{\alpha}$ is $acx^2 \, + \, b(a \, + \, c) \, x \, + \, (a \, + \, c)^2 \, = \, 0$

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

  1. True

  2. False

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Correct Option: A
Explanation:

$\Rightarrow$  $\alpha$ and $\beta$ are roots of the equation $ax^2+bx+c=0$

$\Rightarrow$  $\alpha+\beta=\dfrac{-b}{a}$                       -------- ( 1 )
$\Rightarrow$  $\alpha\beta=\dfrac{c}{a}$                    ------- ( 2 )
Now,
$\Rightarrow$  $\alpha+\dfrac{1}{\beta}+\beta+\dfrac{1}{\alpha}=(\alpha+\beta)+\left(\dfrac{1}{\beta}+\dfrac{1}{\alpha}\right)$

                                    $=(\alpha+\beta)+\left(\dfrac{\alpha+\beta}{\alpha\beta}\right)$
                 
                                    $=\dfrac{-b}{a}+\dfrac{\dfrac{-b}{a}}{\dfrac{c}{a}}$     [ By using ( 1 ) and ( 2 ) ]

                                    $=\dfrac{-b}{a}-\dfrac{b}{c}$

                                    $=\dfrac{-bc-ba}{ac}$

$\therefore$   $\alpha+\dfrac{1}{\beta}+\beta+\dfrac{1}{\alpha}=\dfrac{-b(a+c)}{ac}$                    ----- ( 3 )

$\Rightarrow$  $\left(\alpha+\dfrac{1}{\beta}\right)\left(\beta+\dfrac{1}{\alpha}\right)=\alpha\beta+1+1+\dfrac{1}{\alpha\beta}$
 
                                            $=\dfrac{c}{a}+2+\dfrac{1}{\dfrac{c}{a}}$

                                            $=\dfrac{c}{a}+2+\dfrac{a}{c}$
 
                                            $=\dfrac{a^2+2ac+c^2}{ac}$

$\therefore$  $\left(\alpha+\dfrac{1}{\beta}\right)\left(\beta+\dfrac{1}{\alpha}\right)=\dfrac{a^2+2ac+c^2}{ac}$              ----- ( 4 )
Now, new equation,

$\Rightarrow$  $x^2-\left(\alpha+\dfrac{1}{\beta}+\beta+\dfrac{1}{\alpha}\right)x+\left[\left(\alpha+\dfrac{1}{\beta}\right)\left(\beta+\dfrac{1}{\alpha}\right)\right]=0$
By using ( 3 ) and ( 4 ),

$\Rightarrow$  $x^2+\dfrac{b(a+c)}{ac}x+\dfrac{a^2+2ac+c^2}{ac}$

$\Rightarrow$  $acx^2+b(a+c)x+(a+c)^2=0$

If $\dfrac{x^2 - bx}{ax - c} = \dfrac{m - 1}{m + 1}$ has roots which are numerically equal but of opposite sings, the value of m must be:

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

  1. $\dfrac{a-b}{a + b}$

  2. $\dfrac{a + b}{a - b}$

  3. c

  4. $\dfrac{1}{c}$

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Correct Option: A
Explanation:

$\Longrightarrow \cfrac { { x }^{ 2 }-bx }{ ax-c } =\cfrac { m-1 }{ m+1 } \ \Longrightarrow (m+1){ x }^{ 2 }-b(m+1)x=(m-1)ax-c(m-1)\ \Longrightarrow (m+1){ x }^{ 2 }-[b(m+1)+(m-1)a]x+c(m-1)=0$

Roots are numerically equal but of opposite sign.
$\therefore$ Sum of roots = 0
$\Longrightarrow (b+a)m+(b-a)=0$ 
$\therefore m=\cfrac { a-b }{ a+b } $

If $\alpha$ and $\beta$ are the roots of the equation $ax^{2} \, + \, bx \, + \, c \, = \, 0$. The equation whose roots are as given below.
$\dfrac{\alpha }{\beta } \, ,\dfrac{\beta }{\alpha}$ is $acx^2 \, - \, (b^2 \, - \, 2ac) \, x \, + \, ac \, = \, 0$

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:

$\Rightarrow$  $\alpha$ and $\beta$ are roots of the equation $ax^2+bx+c=0$

$\Rightarrow$  $\alpha\beta=\dfrac{c}{a}$           ------ ( 1 )
$\Rightarrow$  $\alpha+\beta=\dfrac{-b}{a}$         ----- ( 2 )
$\Rightarrow$  $(\alpha+\beta)^2=\alpha^2+\beta^2+2\alpha\beta$
Using ( 1 ) and ( 2 ),
$\Rightarrow$  $\left(\dfrac{-b}{a}\right)^2=\alpha^2+\beta^2+2\times \dfrac{c}{a}$

$\Rightarrow$  $\dfrac{b^2}{a^2}=\alpha^2+\beta^2+\dfrac{2c}{a}$

$\therefore$   $\alpha^2+\beta^2=\dfrac{b^2}{a^2}-\dfrac{2c}{a}$

$\therefore$  $\alpha^2+\beta^2=\dfrac{b^2-2ac}{a^2}$          ------ ( 3 )
Now,
$\Rightarrow$  $\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}=\dfrac{\alpha^2+\beta^2}{\alpha\beta}$

                     $=\dfrac{\dfrac{b^2-2ac}{a^2}}{\dfrac{c}{a}}$           [ Using ( 1 ) and ( 3 ) ]

$\Rightarrow$  $\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}=\dfrac{b^2-2ac}{ac}$           ----- ( 4 )

$\Rightarrow$  $\dfrac{\alpha}{\beta}.\dfrac{\beta}{\alpha}=1$   ---- ( 5 )
New equation,
$\Rightarrow$  $x^2-\left(\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}\right)x+\left(\dfrac{\alpha}{\beta}.\dfrac{\beta}{\alpha}\right)=0$
By Using ( 4 ) and ( 5 ),
$\Rightarrow$  $x^2-\left(\dfrac{b^2-4ac}{ac}\right)x+1=0$

$\Rightarrow$  $acx^2-(b^2-4ac)x+1=0$
$\therefore$  We can see equation given in question is correct.

A quadratic polynomial $p(x)$ with $3$ and $\dfrac{-2}{5}$  as the sum and product of zeroes, respectively is $10x^2+30x-4$

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: B
Explanation:

$10x^2+30x-4=0$

$\alpha+\beta=-\cfrac{30}{10}=-3$
$\alpha\beta=\cfrac{-4}{10}=\cfrac{-2}{5}$
So, a quadratic polynomial $p(x)$ with $3$ and $\cfrac{-2}{5}$ as the sum and product of zeroes, respectively is $10x^2+30x-4$ is false.

If the roots of a quadratic equation are reciprocals of the roots of $ax^2 + bx + c = 0$, then what will be the coefficient of $c$?

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

  1. $x$

  2. $x^2$

  3. $-x$

  4. $x^3$

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Correct Option: B
Explanation:

We know to find the equation with reciprocal roots, we simply put $\dfrac {1}{x}$ in the place of $x$.
so, $a(\dfrac {1}{x})^2 + b(\dfrac {1}{x}) + c = 0$
On taking LCM we get the required equation
$a + bx + cx^2 = 0$
So the coefficient of $c$ is $x^2$

Find the Quadratic Equation whose roots are Reciprocal of $ax^2 + bx + c = 0$.

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

  1. $ax^2 + bx + c = 0$

  2. $a(\dfrac {1}{x})^2 + b(\dfrac {1}{x}) + c = 0$

  3. $3ax^2 + 2bx + c = 0$

  4. None of the above

Show answer
Correct Option: B
Explanation:

We know that for reciprocal roots, we only need to replace $x$ by $\dfrac {1}{x}$, in the given equation.
So the above equation becomes:
$a(\dfrac {1}{x})^2 + b(\dfrac {1}{x}) + c = 0$, which is the required answer.

If A.M. of the roots of a quadratic equation is $8/5$ and A.M. of their reciprocals is $8/7$, then the equation is?

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

  1. $7x^2-16x+8=0$

  2. $3x^2-12x+7=0$

  3. $5x^2-16x+7=0$

  4. $7x^2-16x+5=0$

Show answer
Correct Option: C
Explanation:

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

According to the question,
$\Rightarrow$ $\dfrac{\alpha+\beta}{2}=\dfrac{8}{5}$

$\therefore$  $\alpha+\beta=\dfrac{16}{5}$                ----- ( 1 )

According to the equation,
$\Rightarrow$  $\dfrac{\dfrac{1}{\alpha}+\dfrac{1}{\beta}}{2}=\dfrac{8}{7}$
$\Rightarrow$  $\dfrac{\alpha+\beta}{2\alpha\beta}=\dfrac{8}{7}$
$\Rightarrow$  $7(\alpha+\beta)=16\alpha\beta$
$\Rightarrow$  $\dfrac{7\times \dfrac{16}{5}}{16}=\alpha\beta$
$\therefore$    $\alpha\beta=\dfrac{7}{5}$          - ------ ( 2 )
Now, new eqution,
$\Rightarrow$  $x^2-(\alpha+\beta)x+(\alpha.\beta)=0$
From ( 1 ) and ( 2 ),
$\Rightarrow$  $x^2-\dfrac{16}{5}x+\dfrac{7}{5}=0$
$\Rightarrow$  $5x^2-16x+7=0$   

If $\alpha, \beta$ are the root of a quadratic equation $x^2 - 3x+5=0$, then the equation whose roots are $(\alpha^2 - 3 \alpha +7)$ and $(\beta^2 -3\beta +7)$ is

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

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

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

  3. $x^2 -4x-1=0$

  4. $x^2 +2x+3=0$

Show answer
Correct Option: B
Explanation:

Since $\alpha, \beta$ are the root of equation $x^2-3x+5=0$
So, $\alpha^2-3\alpha +5=0$
$\beta^2 -3\beta +5=0$
$\therefore \alpha^2 -3\alpha =-5$
$\beta^2 - 3 \beta =-5$
Putting in $(\alpha^2 - 3 \alpha +7) $  &  $(\beta^2 - 3\beta +7)$    ....... (1)
$-5 +7, -5 +7$
$\therefore$ 2 and 2 are the roots
$\therefore$ The required equation is $x^2 - 4x+4=0$