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)

$\huge\orange{\overbrace{\red{\underbrace\color{blue}{\underbrace\color{black}{\colorbox{lime}{{\red\:{「Answer」}}}}}}}}$

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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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нσρє ıт нєłρs yσυ

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тнαηkyσυ

Answer 2

${(2x-1) (x - 3) = (x + 5) (x-1)}</p><p></p><p>\huge\orange{\overbrace{\red{\underbrace\color{blue}{\underbrace\color{black}{\colorbox{lime}{{\red\:{「Answer」}}}}}}}} </p><p>「Answer」</p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p></p><p>_______________________________________</p><p></p><p>_______________________________________</p><p></p><p>= > (2x-1) (x - 3) = (x + 5) (x-1)=>(2x−1)(x−3)=(x+5)(x−1)</p><p></p><p>= > 2x(x - 3) - 1(x - 3) = x(x - 1) + 5(x - 1)=>2x(x−3)−1(x−3)=x(x−1)+5(x−1)</p><p></p><p>= > 2 {x}^{2} - 6x - x + 3 = {x}^{2} - x + 5x - 5=>2x </p><p>2</p><p> −6x−x+3=x </p><p>2</p><p> −x+5x−5</p><p></p><p>= > 2 {x}^{2} - {x}^{2} - 6x - x + x - 5x = - 5 - 3=>2x </p><p>2</p><p> −x </p><p>2</p><p> −6x−x+x−5x=−5−3</p><p></p><p>= > {x}^{2} - 6x - 5x = - 8=>x </p><p>2</p><p> −6x−5x=−8</p><p></p><p>= > {x}^{2} - 11x + 8 = 0=>x </p><p>2</p><p> −11x+8=0</p><p></p><p>Here we use quadratic formula to find out the solution:-</p><p></p><p>Here,a=1 ; b=-11 & c=8</p><p></p><p>= > x = \frac{ - b± \sqrt{ {b}^{2} - 4ac} }{2a}=>x= </p><p>2a</p><p>−b± </p><p>b </p><p>2</p><p> −4ac</p><p> </p><p> </p><p> </p><p> </p><p></p><p>= > x = \frac{ - ( - 11)± \sqrt{ {( - 11)}^{2} - 4(8)} }{2}=>x= </p><p>2</p><p>−(−11)± </p><p>(−11) </p><p>2</p><p> −4(8)</p><p> </p><p> </p><p> </p><p> </p><p></p><p>= > x = \frac{11± \sqrt{121 - 32} }{2}=>x= </p><p>2</p><p>11± </p><p>121−32</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\bold{\red{= > x = \frac{11± \sqrt{89} }{2}}}=>x= </p><p>2</p><p>11± </p><p>89</p><p> </p><p> </p><p> </p><p> </p><p></p><p>_______________________________________</p><p></p><p>_______________________________________</p><p></p><p>нσρє ıт нєłρs yσυ</p><p></p><p>_____________________</p><p></p><p> \: тнαηkyσυ$