Tag: forming quadratic equation

Questions Related to forming quadratic equation

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

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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

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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

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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.

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

Show answer
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. $\dfrac{1}{c}$

Show answer
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 } $