Question

The equation whose roots are multiplied by 3 of those of 2x^2 + 3x – 1=0 is​

Answer 1

$\text{Let the roots of 2x^2+3x-1=0 be \alpha and \beta }$

$\textbf{To find:}$

$\text{The quadratic equation whose roots are 3\alpha and 3\beta }$

$\textbf{Solution:}$

$\text{Consider,}$

$2x^2+3x-1=0$

$\text{Sum of the roots}=\dfrac{-3}{2}$

$\implies\bf\alpha+\beta=\dfrac{-3}{2}$

$\text{Product of the roots}=\dfrac{-1}{2}$

$\implies\bf\alpha\,\beta=\dfrac{-1}{2}$

$\text{Now, we form a quadratic whose roots are 3\alpha and 3\beta }$

$3\alpha+3\beta=3(\alpha+\beta)=3(\dfrac{-3}{2})=\dfrac{-9}{2}$

$3\alpha{\times}3\beta=9(\alpha,\beta)=9(\dfrac{-1}{2})=\dfrac{-9}{2}$

$\textbf{The required equation is}$

$x^2-(3\alpha+3\beta)x+(3\alpha\,3\beta)=0$

$x^2-(\dfrac{-9}{2})x+(\dfrac{-9}{2})=0$

$x^2+\dfrac{9}{2}x-\dfrac{9}{2}=0$

$\implies\boxed{\bf\,2x^2+9x-9=0}$

Answer 2

Answer:

Let the roots of 2x

2

+3x−1=0 be α and β

\textbf{To find:}To find:

\text{The quadratic equation whose roots are $3\alpha$ and $3\beta$}The quadratic equation whose roots are 3α and 3β

\textbf{Solution:}Solution:

\text{Consider,}Consider,

2x^2+3x-1=02x

2

+3x−1=0

\text{Sum of the roots}=\dfrac{-3}{2}Sum of the roots=

2

−3

\implies\bf\alpha+\beta=\dfrac{-3}{2}⟹α+β=

2

−3

\text{Product of the roots}=\dfrac{-1}{2}Product of the roots=

2

−1

\implies\bf\alpha\,\beta=\dfrac{-1}{2}⟹αβ=

2

−1

\text{Now, we form a quadratic whose roots are $3\alpha$ and $3\beta$}Now, we form a quadratic whose roots are 3α and 3β

3\alpha+3\beta=3(\alpha+\beta)=3(\dfrac{-3}{2})=\dfrac{-9}{2}3α+3β=3(α+β)=3(

2

−3

)=

2

−9

3\alpha{\times}3\beta=9(\alpha,\beta)=9(\dfrac{-1}{2})=\dfrac{-9}{2}3α×3β=9(α,β)=9(

2

−1

)=

2

−9

\textbf{The required equation is}The required equation is

x^2-(3\alpha+3\beta)x+(3\alpha\,3\beta)=0x

2

−(3α+3β)x+(3α3β)=0

x^2-(\dfrac{-9}{2})x+(\dfrac{-9}{2})=0x

2

−(

2

−9

)x+(

2

−9

)=0

x^2+\dfrac{9}{2}x-\dfrac{9}{2}=0x

2

+

2

9

x−

2

9

=0

\implies\boxed{\bf\,2x^2+9x-9=0}⟹

2x

2

+9x−9=0