Question

(2x-1) (x - 3) = (x + 5) (x-1)​

Answer 1

$\huge\tt{\bold{\underline{\underline{Question᎓}}}}$

(2x-1) (x - 3) = (x + 5) (x-1)

$\huge\tt{\bold{\underline{\underline{Answer᎓}}}}$

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$= > (2x-1) (x - 3) = (x + 5) (x-1)$

$= > 2x(x - 3) - 1(x - 3) = x(x - 1) + 5(x - 1)$

$= > 2 {x}^{2} - 6x - x + 3 = {x}^{2} - x + 5x - 5$

$= > 2 {x}^{2} - {x}^{2} - 6x - x + x - 5x = - 5 - 3$

$= > {x}^{2} - 6x - 5x = - 8$

$= > {x}^{2} - 11x + 8 = 0$

Here we use quadratic formula to find out the solution:-

Here,a=1 ; b=-11 & c=8

$= > x = \frac{ - b± \sqrt{ {b}^{2} - 4ac} }{2a}$

$= > x = \frac{ - ( - 11)± \sqrt{ {( - 11)}^{2} - 4(8)} }{2}$

$= > x = \frac{11± \sqrt{121 - 32} }{2}$

$\bold{\red{= > x = \frac{11± \sqrt{89} }{2}}}$

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Answer 2

$\bold{\underline{ \Huge\sf \color{grey}{GIVEN\dag}}}$

$\large\tt(2x-1) (x - 3) = (x + 5) (x-1)$

$\bold{\underline {\Huge{\displaystyle\sf \gray{TO \: FIND \dag}}}}$

$\bold\bull \: \tt{The \: value \: of \: x}$

$\bold{\underline{\Huge{\displaystyle\sf \gray{SOLUTION \dag}}}}$

$\implies\bf(2x-1) (x - 3) = (x + 5) (x-1)$

$\implies\bf2x(x - 3) - 1(x - 3) = x(x - 1) + 5(x - 1)$

$\implies\bf2 {x}^{2} - 6x - x + 3 = {x}^{2} - x + 5x - 5$

$\implies\bf2 {x}^{2} - {x}^{2} - 6x - x + x - 5x = - 5 - 3$

$\implies\bf{x}^{2} - 6x - 5x = - 8$

$\implies \bf{x}^{2} - 11x + 8 = 0$

Here we use quadratic formula to find out the solution:-

Here,a=1 ; b=-11 & c=8

$\implies \bf{x = \frac{ - b± \sqrt{ {b}^{2} - 4ac} }{2a}}$

$\implies \bf{x = \frac{ - ( - 11)± \sqrt{ {( - 11)}^{2} - 4(8)} }{2}}$

$\implies\bf{x = \frac{11± \sqrt{121 - 32} }{2}}$

$\implies\bold{\color{fuchsia}{ x = \frac{11± \sqrt{89} }{2}}}$