Vieta’s formula for quadratic equations - class-XI

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

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

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

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.

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

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

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

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

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 $x^2$

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.

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

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

$a _1x^2+b _1x+c _1=0$

$\Rightarrow$  $\alpha+\beta=-2$             ------ ( 1 )

Let the roots of the equation be a and b
then, $a^3 - b^3 = 189$
and $a - b = 3$
cubing both sides:
$(a-b)^3 = 27 $
$a^3 - b^3 - 3ab (a-b) = 27$
$189 -3ab(3) = 27 $
$162 = 9 ab$
$ab = 18$
Similarly, $(a+b)^2 = (a -b)^2  + 4ab$
$(a+b)^2 = 3^2 + 4(18)$
$(a+b)^2 = 9 + 72 $
$a +b = \pm 9$
The general form of equation is $x^2 -Sx + P = 0 $, hence the equation will be
$x^2 \pm 9x + 18 = 0$

If $\alpha, \beta$ are the roots of $x^2-2x+3=0$
then $\displaystyle \alpha ^{2}-2\alpha +3= 0$ ...(1)
and $\displaystyle \beta^2-2\beta+3=0$  ....(2)
$\displaystyle \therefore \alpha ^{2}= 2\alpha -3, \alpha ^{3}= 2\alpha ^{2}-3\alpha $
$\displaystyle \therefore P= \left ( 2\alpha ^{2}-3\alpha  \right )-3\alpha ^{2}+5\alpha -2$
$\displaystyle = -\alpha ^{2}+2\alpha -2= 3-2= 1,$ by (1)
Similarly $\displaystyle Q= 2 \therefore S= 3, P= 2$
Hence reqd. eq. is $\displaystyle x^{2}-3x+2= 0.$ or $-x^2+3x-2=0$

Let the roots of the equation be a and b
then, $a^3 - b^3 = 208$
and $a - b = 4$
cubing both sides:
$(a-b)^3 = 64 $
$a^3 - b^3 - 3ab (a-b) = 64$
$208 -3ab(4) = 64 $
$144 = 12 ab$
$ab = 12$
Similarly, $(a+b)^2 = (a -b)^2  + 4ab$
$(a+b)^2 = 4^2 + 4(12)$
$(a+b)^2 = 16 + 48 $
$a +b = \pm 8$
The general form of equation is $x^2 -Sx + P = 0 $, hence the equation will be
$x^2 \pm 8x + 12 = 0$

$ { x }^{ 2 }+x+1=0$

Let $\alpha$ and $\beta$ be the roots of the equation.
Hence
$\alpha+\beta=4$
and $\alpha^{3}+\beta^{3}=28$
Now $\alpha^{3}+\beta^{3}$ can be written as

Let one root of $ax^2 + bx + c =0$ be $\alpha$ and other be $\alpha^2$
then, $\alpha + \alpha^2 = \frac{-b}{a}$
$\alpha^3 = \frac{c}{a}$
or $\alpha = (\frac{c}{a})^{({\frac{1}{3}})}$
Now put the value of $\alpha$ in $\alpha + \alpha^2 = \frac{-b}{a}$
$\frac{c}{a}^{\frac{1}{3}} + \frac{c}{a}^{\frac{2}{3}} = \frac{-b}{a}$
Cubing both sides:

Let $\alpha ,\beta $ are roots of $2{ x }^{ 2 }-3x+5=0$ 
Then to get equation whose roots are $\displaystyle \dfrac { 1 }{ \alpha  } ,\dfrac { 1 }{ \beta  } $ 
Replace $\displaystyle x\rightarrow \frac { 1 }{ x } \ $
We get $5{ x }^{ 2 }-3x+2=0$.

$ The\quad roots\quad of\quad the\quad equation\quad { x }^{ 2 }+11x+13=0\quad is\quad diminished\quad by\quad 4\quad i.e.\ the\quad variable\quad becomes\quad (x+2).\ \therefore \quad
The\quad new\quad equation\quad is\quad \ (x+4)^{ 2 }+11(x+4)+13=0\ \Rightarrow { x }^{ 2 }+8x+16+11x+44+13=0\ \Rightarrow { x }^{ 2 }+19x+73=0\ Ans-\quad Option\quad C .$

For the equation, sum of roots $ = \alpha  + \beta = -\dfrac {3}{1} = -3 $
Product of roots $ = \alpha  \times \beta = \dfrac {3}{1} = 3 $

So, the eqn with roots $ = \alpha  + \beta$ and $ \alpha  \times \beta $ is $ (x - (-3))(x-3) = 0 $
$ => x^{2} -9 = 0 $

The equation can be written as $x^{ 2 }+px+q=0$

$x^2 -3x = 3 = 0$ ........ (1)
Here, $a +\beta =3$ and $a\beta =3$. Therefore,
$2(a +\beta ) =6$
$2\times 2a\beta = 4a\beta =4 \times 3= 12$
The equation whose roots are double of (1),
$x^2 -$(sum of the roots)x + product of the roots $=0$
will be $x^2 -6x + 12 =0$

The given equation is $x^2 -x + 1 = 0$ ....... (1)
Here, $a +\beta = 1$ and $a\beta = 1$. Therefore,
$a^2+\beta^2 = (a + \beta)^2 -2a\beta = 1-2= -1$
and $a^2\beta^2 = (a\beta)^2 = 1^2= 0$
Therefore, the equation whose roots are square of 1 is $x^2$ -(sum of the roots)x +product $=0$
or $x^2-(-1)x+ 1 =0$
or $x^2+x+ 1 =0$

$(x+p)(x+q)-k=0\ \Longrightarrow { x }^{ 2 }+(p+q)x+pq-k=0$

${ x }^{ 2 }+p=0$

let $p,q$ be roots of equation $ax^2+bx+c=0$

From the first equation, we can conclude that
$\alpha+\beta=-\cfrac{b}a$  ...(i)

$2x^{2}-3x-6=0$

$\alpha, \beta$ are the roots of $x^2 + px+1=0$
$\Rightarrow \alpha+\beta = -p,  \alpha \beta =1$

In the given equation
sum of roots $= -5$ and product of roots $= 6$
The standard form of a quadratic equation is: $x^2 - (S)x + P = 0$, where S and P are sum and product of roots.
So according to question
$x^2 - (6)x + (-5) = 0$
The required equation will be $x^2 - 6x - 5 = 0$

$\alpha, \beta$ are the roots of $\Rightarrow { ax }^{ 2 }+bx+c=0$
Then, $a(\alpha)^2 + b(\alpha)+c =0$
Now, $\alpha +2 = x $
Hence, $\alpha = x -2$
Thus, replace $\alpha$ by $x-2$ in the given equation,
Required equation is
$a(x-2)^2+b(x-2)+c=0$
$\Rightarrow a(x^2-4x+4)+bx-2b+c=0$
$\Rightarrow ax^2 +(b-4a) x+(4a-2b+c)=0$
$\Rightarrow ax^{ 2 }+x(b-4a)+4a-2b+c=0$

We have, $ax^2-bx^2+bx+cx^2-2cx+c=0$

Since $\alpha$ and $\beta$ are roots of the equation
$x^{2}+px+q = 0$, therefore
$\alpha + \beta = -p$      ...(i)
and $\alpha\beta = q$      ...(ii)
Sum of the roots $= \alpha^{2}+\alpha\beta+\beta^{2}+\alpha\beta$
$= (\alpha+\beta)^{2} = p^{2}$
Product of the roots $=(\alpha^{2}+\alpha\beta)(\beta^{2}+\alpha\beta)$
$= \alpha\beta (\alpha+\beta)^{2} = qp^{2}$
Required equation will be
$x^{2}$-(Sum  of  the  roots)$x$ + Product  of  the  roots = $0$
or $x^{2}-p^{2}x+qp^{2} = 0$

$\displaystyle \alpha +\beta = 2$ and let $\displaystyle \alpha \beta = p$
$\displaystyle \therefore $ Equation is $\displaystyle x^{2}-2x+p= 0$ ...(1)
We have to find the value of p.
Now $\displaystyle

\frac{1-\alpha }{1+\beta }+\frac{1-\beta }{1+\alpha }= \frac{\left (

1-\alpha ^{2} \right )+\left ( 1-\beta ^{2} \right )}{1+\left ( \alpha

+\beta  \right )+p}$
or $\displaystyle \frac{2-\left ( \alpha

^{2}+\beta ^{2} \right )}{1+2+p}= \frac{2-\left { \left ( \alpha +\beta

 \right )^{2}-2\alpha \beta  \right }}{3+p}$
or $\displaystyle

\frac{2-4+2p}{3+p}:or:\frac{2\left ( p-1 \right )}{p+3}= 2\left (

\frac{4\lambda ^{2}+15}{4\lambda ^{2}-1} \right )$
or $\displaystyle \frac{p-1}{p+3}= \frac{4\lambda ^{2}+15}{4\lambda ^{2}-1}$
or $\displaystyle

p\left [ \left ( 4\lambda ^{2}-1 \right )-\left ( 4\lambda ^{2}+15

\right ) \right ]= 3\left ( 4\lambda ^{2}+15 \right )+\left ( 4\lambda

^{2}-1 \right )$
or $\displaystyle -16p= 16\lambda ^{2}+44= 4\left ( 4\lambda ^{2}+11 \right )$
$\displaystyle \therefore p= -\frac{\left ( 4\lambda ^{2}+11 \right )}{4}$
Putting for $p$ in (1) we get the required equation as
$\displaystyle x^{2}-2x-\frac{\left ( 4\lambda ^{2}+11 \right )}{4}= 0$

For the given equation, sum of roots $ = \alpha + \beta  = -\dfrac {1}{1} = -1 $

Product of roots $ = \alpha \times \beta =  \dfrac {1}{1} =1 $

Now, $ {\alpha}^{2} + \beta ^{2} = (\alpha + \beta )^{2} - 2(\alpha \times \beta)= (-1)^{2} - 2(1) = 1-2 = -1 $

And $ {\alpha}^{2} \times \beta ^{2} = (\alpha \times \beta )^{2} = 1 $

Equation whose roots are $ {\alpha}^{2} $ and $ \beta ^{2} $ is $ x^{2} -(Sum \ of \ roots)x +  Product \ of \ roots  = 0 $
$ => x^{2} -({\alpha}^{2} + \beta ^{2})x +  {\alpha}^{2} \times \beta ^{2}  = 0 $
$ => x^{2} -(-1)x+ 1  = 0 $
$ => x^{2} +x+ 1  = 0 $

Given: Ragini found roots as $3, -\dfrac 12$ when copied the wrong coefficient of $x$ and Gourav found roots as $-1, -\dfrac 14$ when copied wrong constant term.

With the roots $ 4, 12 $, the equation was $ (x-4))(x-12) ={x}^{2} -4x  -12x + 48 = {x}^{2} -16x + 48 $

As
Rohan made a mistake in noting the coffecient of $ x $ , in the original
equation, coefficient of $ {x}^{2} = 1 $ and constant $ = 48 $

Now,
with the roots $ -19, 3 $, the equation was $ (x-(-19))(x-3) =
(x+19)(x-3) = {x}^{2} + 19x -3x -57 = {x}^{2} + 16x -57 $

As Sohan made a mistake in noting just the constant term in the original 
equation, coefficient of $ {x}^{2} = 1 $ and of $ x = 16 $

So, we get the original equation as $ {x}^{2} + 16x + 48 = 0 $
Solving it, we get $ {x}^{2} + 4x +12x + 48 = 0 $
$ => x(x+4) + 12(x+4) = 0 $
$ => (x+4)(x+12) = 0 $
$ => x = -4, -12 $

Since the equation $2{ x }^{ 2 }+8x+2=0$ has roots which are 1 less than those of the equation $a{ x }^{ 2 }+bx+c=0$ then if we replace $ x $ by $ x+1 $ in latter we'll get the former.
$\Rightarrow a(x+1)^{ 2 }+b(x+1)+c=0$
$\Rightarrow  a{ x }^{ 2 }+(2a+b)x+a+b+c=0$
comparing this equation with that of $2{ x }^{ 2 }+8x+2=0$
we get option (b) as the correct answer

With the roots $ -3, -12 $, the equation was $ (x-(-3))(x-(-12) = (x+3)(x+6) = {x}^{2} + 3x + 12x + 36 = {x}^{2} + 15x + 36 $

As Umesh made a mistake in noting just the constant term, in the original equation, coefficient of $ {x}^{2} = 1 $ and of $ x = 15 $

Now, with the roots $ -27, -2 $, the equation was $ (x-(-27))(x-(-2)) = (x+27)(x+2) = {x}^{2} + 27x + 2x + 54 = {x}^{2} + 29x + 54 $

As Varun made a mistake in noting the coffecient of $ x $ in the original 
equation, coefficient of $ {x}^{2} = 1 $ and constant $ = 54 $

So, we get the original equation as $ {x}^{2} + 15x + 54 = 0 $