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# If alpha and beta are the zeros pf p(x)= 3x^2+5x-2 then form a quadratic polynomial whose zeros are 2alpha and 2beta

Shreya D Nanda
74 Points
4 years ago
Hi Riya,
If $\dpi{80} \alpha , and \beta$ are roots of a quadratic equation, say, $\dpi{80} f\left ( x \right )=ax^2+ bx + c$, then, we have
$\dpi{80} \alpha +\beta = -b/a$  and $\dpi{80} \alpha\beta = c/a$
$\dpi{80} p\left ( x \right ) =3x^2 +5x-2$
$\dpi{80} \therefore \alpha +\beta =-5/3$  and   $\dpi{80} \therefore \alpha\beta =-2/3$

To form an another equation, where $\dpi{80} 2\alpha , 2\beta$ are the roots then,
$\dpi{80} 2\alpha + 2\beta= 2\left ( \alpha +\beta \right ) =2*-5/3=-10/3$,   say      $\dpi{80} -b'/a'=-10/3 \Rightarrow a'=3 , b'=10$
$\dpi{80} [2\alpha][2\beta] =4\alpha \beta =4*-2/3= -8/3$    , say    $\dpi{80} c''/a'=-8/3 \Rightarrow a'=3 , c'=-8$

Therefore,  the new quadratic polynomial, with $\dpi{80} 2\alpha , 2\beta$ as roots, is
$\dpi{80} p'\left ( x \right ) =a'x^2+b'x +c'$
$\dpi{80} \Rightarrow p'\left ( x \right ) =3x^2+10x -8'$.

hope it is clear,,
sachin
85 Points
4 years ago

If $\dpi{80} \alpha , and \beta$ are roots of a quadratic equation, say, $\dpi{80} f\left ( x \right )=ax^2+ bx + c$, then, we have
$\dpi{80} \alpha +\beta = -b/a$  and $\dpi{80} \alpha\beta = c/a$
$\dpi{80} p\left ( x \right ) =3x^2 +5x-2$
$\dpi{80} \therefore \alpha +\beta =-5/3$  and   $\dpi{80} \therefore \alpha\beta =-2/3$

To form an another equation, where $\dpi{80} 2\alpha , 2\beta$ are the roots then,
$\dpi{80} 2\alpha + 2\beta= 2\left ( \alpha +\beta \right ) =2*-5/3=-10/3$,   say      $\dpi{80} -b'/a'=-10/3 \Rightarrow a'=3 , b'=10$
$\dpi{80} [2\alpha][2\beta] =4\alpha \beta =4*-2/3= -8/3$    , say    $\dpi{80} c''/a'=-8/3 \Rightarrow a'=3 , c'=-8$

Therefore,  the new quadratic polynomial, with $\dpi{80} 2\alpha , 2\beta$ as roots, is
$\dpi{80} p'\left ( x \right ) =a'x^2+b'x +c'$
$\dpi{80} \Rightarrow p'\left ( x \right ) =3x^2+10x -8'$.
sachin
85 Points
4 years ago

If $\dpi{80} \alpha , and \beta$ are roots of a quadratic equation, say, $\dpi{80} f\left ( x \right )=ax^2+ bx + c$, then, we have
$\dpi{80} \alpha +\beta = -b/a$  and $\dpi{80} \alpha\beta = c/a$
$\dpi{80} p\left ( x \right ) =3x^2 +5x-2$
$\dpi{80} \therefore \alpha +\beta =-5/3$  and   $\dpi{80} \therefore \alpha\beta =-2/3$

To form an another equation, where $\dpi{80} 2\alpha , 2\beta$ are the roots then,
$\dpi{80} 2\alpha + 2\beta= 2\left ( \alpha +\beta \right ) =2*-5/3=-10/3$,   say      $\dpi{80} -b'/a'=-10/3 \Rightarrow a'=3 , b'=10$
$\dpi{80} [2\alpha][2\beta] =4\alpha \beta =4*-2/3= -8/3$    , say    $\dpi{80} c''/a'=-8/3 \Rightarrow a'=3 , c'=-8$

Therefore,  the new quadratic polynomial, with $\dpi{80} 2\alpha , 2\beta$ as roots, is
$\dpi{80} p'\left ( x \right ) =a'x^2+b'x +c'$.
$\dpi{80} \Rightarrow p'\left ( x \right ) =3x^2+10x -8'$.
sachin
85 Points
4 years ago

If $\dpi{80} \alpha , and \beta$ are roots of a quadratic equation, say, $\dpi{80} f\left ( x \right )=ax^2+ bx + c$, then, we have
$\dpi{80} \alpha +\beta = -b/a$  and $\dpi{80} \alpha\beta = c/a$
$\dpi{80} p\left ( x \right ) =3x^2 +5x-2$
$\dpi{80} \therefore \alpha +\beta =-5/3$  and   $\dpi{80} \therefore \alpha\beta =-2/3$

To form an another equation, where $\dpi{80} 2\alpha , 2\beta$ are the roots then,
$\dpi{80} 2\alpha + 2\beta= 2\left ( \alpha +\beta \right ) =2*-5/3=-10/3$,   say      $\dpi{80} -b'/a'=-10/3 \Rightarrow a'=3 , b'=10$
$\dpi{80} [2\alpha][2\beta] =4\alpha \beta =4*-2/3= -8/3$    , say    $\dpi{80} c''/a'=-8/3 \Rightarrow a'=3 , c'=-8$

Therefore,  the new quadratic polynomial, with $\dpi{80} 2\alpha , 2\beta$ as roots, is
$\dpi{80} p'\left ( x \right ) =a'x^2+b'x +c'$
$\dpi{80} \Rightarrow p'\left ( x \right ) =3x^2+10x -8'$…,.,.
sachin
85 Points
4 years ago

If $\dpi{80} \alpha , and \beta$ are roots of a quadratic equation, say, $\dpi{80} f\left ( x \right )=ax^2+ bx + c$, then, we have
$\dpi{80} \alpha +\beta = -b/a$  and $\dpi{80} \alpha\beta = c/a$
$\dpi{80} p\left ( x \right ) =3x^2 +5x-2$
$\dpi{80} \therefore \alpha +\beta =-5/3$  and   $\dpi{80} \therefore \alpha\beta =-2/3$

To form an another equation, where $\dpi{80} 2\alpha , 2\beta$ are the roots then,
$\dpi{80} 2\alpha + 2\beta= 2\left ( \alpha +\beta \right ) =2*-5/3=-10/3$,   say      $\dpi{80} -b'/a'=-10/3 \Rightarrow a'=3 , b'=10$
$\dpi{80} [2\alpha][2\beta] =4\alpha \beta =4*-2/3= -8/3$    , say    $\dpi{80} c''/a'=-8/3 \Rightarrow a'=3 , c'=-8$

Therefore,  the new quadratic polynomial, with $\dpi{80} 2\alpha , 2\beta$ as roots, is
$\dpi{80} p'\left ( x \right ) =a'x^2+b'x +c'$
$\dpi{80} \Rightarrow p'\left ( x \right ) =3x^2+10x -8'$.//,/,/,/
sachin
85 Points
4 years ago

If $\dpi{80} \alpha , and \beta$ are roots of a quadratic equation, say, $\dpi{80} f\left ( x \right )=ax^2+ bx + c$, then, we have
$\dpi{80} \alpha +\beta = -b/a$  and $\dpi{80} \alpha\beta = c/a$
$\dpi{80} p\left ( x \right ) =3x^2 +5x-2$
$\dpi{80} \therefore \alpha +\beta =-5/3$  and   $\dpi{80} \therefore \alpha\beta =-2/3$

To form an another equation, where $\dpi{80} 2\alpha , 2\beta$ are the roots then,
$\dpi{80} 2\alpha + 2\beta= 2\left ( \alpha +\beta \right ) =2*-5/3=-10/3$,   say      $\dpi{80} -b'/a'=-10/3 \Rightarrow a'=3 , b'=10$
$\dpi{80} [2\alpha][2\beta] =4\alpha \beta =4*-2/3= -8/3$    , say    $\dpi{80} c''/a'=-8/3 \Rightarrow a'=3 , c'=-8$

Therefore,  the new quadratic polynomial, with $\dpi{80} 2\alpha , 2\beta$ as roots, is
$\dpi{80} p'\left ( x \right ) =a'x^2+b'x +c'$
$\dpi{80} \Rightarrow p'\left ( x \right ) =3x^2+10x -8'$.././/././.....
soumya ranjan prusty
23 Points
4 years ago
If $\dpi{80} \alpha , and \beta$ are roots of a quadratic equation, say, $\dpi{80} f\left ( x \right )=ax^2+ bx + c$, then, we have
$\dpi{80} \alpha +\beta = -b/a$  and $\dpi{80} \alpha\beta = c/a$
$\dpi{80} p\left ( x \right ) =3x^2 +5x-2$
$\dpi{80} \therefore \alpha +\beta =-5/3$  and   $\dpi{80} \therefore \alpha\beta =-2/3$

To form an another equation, where $\dpi{80} 2\alpha , 2\beta$ are the roots then,
$\dpi{80} 2\alpha + 2\beta= 2\left ( \alpha +\beta \right ) =2*-5/3=-10/3$,   say      $\dpi{80} -b'/a'=-10/3 \Rightarrow a'=3 , b'=10$
$\dpi{80} [2\alpha][2\beta] =4\alpha \beta =4*-2/3= -8/3$    , say    $\dpi{80} c''/a'=-8/3 \Rightarrow a'=3 , c'=-8$

Therefore,  the new quadratic polynomial, with $\dpi{80} 2\alpha , 2\beta$ as roots, is
$\dpi{80} p'\left ( x \right ) =a'x^2+b'x +c'$.
$\dpi{80} \Rightarrow p'\left ( x \right ) =3x^2+10x -8'$.
soumya ranjan prusty
23 Points
4 years ago
If $\dpi{80} \alpha , and \beta$ are roots of a quadratic equation, say, $\dpi{80} f\left ( x \right )=ax^2+ bx + c$, then, we have
$\dpi{80} \alpha +\beta = -b/a$  and $\dpi{80} \alpha\beta = c/a$
$\dpi{80} p\left ( x \right ) =3x^2 +5x-2$
$\dpi{80} \therefore \alpha +\beta =-5/3$  and   $\dpi{80} \therefore \alpha\beta =-2/3$

To form an another equation, where $\dpi{80} 2\alpha , 2\beta$ are the roots then,
$\dpi{80} 2\alpha + 2\beta= 2\left ( \alpha +\beta \right ) =2*-5/3=-10/3$,   say      $\dpi{80} -b'/a'=-10/3 \Rightarrow a'=3 , b'=10$
$\dpi{80} [2\alpha][2\beta] =4\alpha \beta =4*-2/3= -8/3$    , say    $\dpi{80} c''/a'=-8/3 \Rightarrow a'=3 , c'=-8$

Therefore,  the new quadratic polynomial, with $\dpi{80} 2\alpha , 2\beta$ as roots, is
$\dpi{80} p'\left ( x \right ) =a'x^2+b'x +c'$.
$\dpi{80} \Rightarrow p'\left ( x \right ) =3x^2+10x -8'$. ok....................................