Question

Solve the following quadratic equations by using formula method. 2n²+5n+2=0​

Answer 1

Answer:

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Answer 2

Given equations:-

$\sf \dfrac{3x}{2} - \dfrac{5y}{3} = - 2..(i)$

$\sf \dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6} ..(ii)$

$\small\underline{\pmb{\sf \:\: \: of \: equation \: 1 :-}}$

$\sf \: \: \: \: \: \: \: \:\pink{ \: \:\::\implies \dfrac{3x}{2} - \dfrac{5y}{3} = - 2}$

$\sf \: \: \: \: \: \: \: \: \: \:\::\implies \dfrac{3(3x) - 2(5y)}{6} = - 2$

$\sf \: \: \: \: \: \: \: \: \: \:\::\implies\dfrac{9x - 10y}{6} = - 2$

$\sf \: \: \: \: \: \: \: \: \: \:\::\implies9x - 10y = - 2(6)$

$\sf \: \: \: \: \: \: \: \: \: \:\::\implies9x - 10y = - 12$

$\sf\: \: \: \: \: \: \: \: \: \:\::\implies 9x = - 12 + 10y$

$\sf \: \: \: \: \: \: \: \: \pink{ \: \:\::\implies x = \dfrac{10y - 12}{9}..(iii)}$

$\sf Substituting \: x = \dfrac{10y - 12}{9} \: in \: equation \: (ii)$

$\qquad\leadsto\quad \sf \dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6}$

$\qquad\leadsto\quad \sf \dfrac{ \bigg(\: \: \: \dfrac{10y - 12}{9 } \bigg) }{3} + \dfrac{y}{2} = \dfrac{13}{6}$

$\qquad\leadsto\quad\sf \dfrac{ 10y - 12 }{27} + \dfrac{y}{2} = \dfrac{13}{6}$

$\qquad\leadsto\quad\sf \dfrac{ 2(10y - 12) + 27y }{54} = \dfrac{13}{6}$

$\qquad\leadsto\quad \sf \dfrac{ 20y - 24 + 27y }{54} = \dfrac{13}{6}$

$\qquad\leadsto\quad \sf 47y - 24 = \dfrac{13}{6} \times 56$

$\qquad\leadsto\quad \sf 47y - 24 = 117$

$\qquad\leadsto\quad \sf 47y = 117 + 24 = 141$

$\qquad\leadsto\quad\red{ \sf y = \cancel{\dfrac{141}{47} }= 3}$

$\sf Substituting \: y = 3\: in \: equation \:(iii)$

$\qquad\leadsto\quad \sf x = \dfrac{10y - 12}{9}$

$\qquad\leadsto\quad \sf x = \dfrac{10(3) - 12}{9}$

$\qquad\leadsto\quad\red{ \sf x = \cancel{\dfrac{18}{9} }= 2}$

$\therefore\:\underline{\textsf{ Value of x and y are \textbf{2 and 3 }}}.$