25n^5-36y^2fl find it please
Answers
Step-by-step explanation:
one roots of the quadratic equation
=> ( a² - 5a + 3 )x² + ( 3a - 2 )x + 2 = 0
=> twice as larger other.
\begin{gathered}\sf : \implies \: let \: we \: assume \: that \: \alpha and \: 2 \alpha \: be \: the \: roots \: of \: equation. \\ \\ \sf : \implies \: sum \: of \: roots \: of \: quadratic \: equation \: = \alpha + \beta = \frac{ - b}{a} \\ \\ \sf : \implies \: \alpha + 2 \alpha = 3 \alpha = \frac{ 1 - 3 a }{ {a}^{2} - 5a + 3} \: \: \: ......(1)\end{gathered}
:⟹letweassumethatαand2αbetherootsofequation.
:⟹sumofrootsofquadraticequation=α+β=
a
−b
:⟹α+2α=3α=
a
2
−5a+3
1−3a
......(1)
\begin{gathered}\sf : \implies \: product \: of \: quadratic \: equation \: = \alpha \beta = \dfrac{c}{a} \\ \\ \sf : \implies \: \alpha (2 \alpha ) = 2 { \alpha }^{2} = \frac{2}{ {a}^{2} - 5a + 3} \: \: \: \: ......(2)\end{gathered}
:⟹productofquadraticequation=αβ=
a
c
:⟹α(2α)=2α
2
=
a
2
−5a+3
2
......(2)
\begin{gathered}\sf : \implies \: on \: squaring \: equation \: (1) \: \: we \: \: get \\ \\ \sf : \implies \: 9 { \alpha }^{2} = \frac{( 1- 3a) {}^{2} }{( {a}^{2} - 5a + 3) {}^{2} } \\ \\ \sf : \implies \: dividing \: equation \: (1) \: \: and \: \: (2) \: \: we \: \: get\end{gathered}
:⟹onsquaringequation(1)weget
:⟹9α
2
=
(a
2
−5a+3)
2
(1−3a)
2
:⟹dividingequation(1)and(2)weget
\begin{gathered}\sf : \implies \: \dfrac{9 { \alpha }^{2} }{2 { \alpha }^{2} } = \dfrac{(1 - 3a) {}^{2} }{( {a}^{2} - 5a + 3) {}^{2} } \times \dfrac{( {a}^{2} - 5a + 3)}{2} \\ \\ \sf : \implies \: \frac{9}{2} = \frac{(1 - 3a) {}^{2} }{2( {a}^{2} - 5a + 3) } \\ \\ \sf : \implies \: 9( {a}^{2} - 5a + 3) = (1 - 3a) {}^{2}\end{gathered}
:⟹
2α
2
9α
2
=
(a
2
−5a+3)
2
(1−3a)
2
×
2
(a
2
−5a+3)
:⟹
2
9
=
2(a
2
−5a+3)
(1−3a)
2
:⟹9(a
2
−5a+3)=(1−3a)
2
\begin{gathered}\sf : \implies \: 9 {a}^{2} - 45a + 27 = 1 + 9 {a}^{2} - 6a \\ \\ \sf : \implies \: 26 = 39a \\ \\ \sf : \implies \: a \: = \frac{2}{3}\end{gathered}
:⟹9a
2
−45a+27=1+9a
2
−6a
:⟹26=39a
:⟹a=
3
2
ok
ok