Question

solve by factorization method​

Answer 1

AnswEr :

(a)$\large\sf\frac{x}{x +1} + \frac{x+1}{x} = \frac{34}{15}$

Let us simplify this first;

$\normalsize\dashrightarrow\sf\frac{x^{2} + \left(x + 1) \right)^{2}}{x \left(x +1 \right)} =\frac{34}{15}$

$\normalsize\dashrightarrow\sf\frac{x^{2} + x^{2} + 2x + 1}{ x^{2} + x} = \frac{34}{15}$

$\quad\scriptsize\dag\sf\ Cross \: multiply \: both \: sides$

$\normalsize\dashrightarrow\sf\ 15 \left(2x^{2} + 2x + 1 \right) = 34 \left(x^{2} + x \right)$

$\normalsize\dashrightarrow\sf\ 30x^{2} + 30x + 15 = 34x^{2} + 34x$

$\normalsize\dashrightarrow\sf\ 30x^{2} + 30x + 15 - 34x^{2} - 34x = 0$

$\normalsize\dashrightarrow\sf\ -4x^{2} - 4x + 15 = 0$

$\quad\scriptsize\dag\sf\ Take \: negative \: sign \: common$

$\normalsize\dashrightarrow\sf\ - \: \left(4x^{2} +4x - 15 \right) = 0$

$\normalsize\dashrightarrow\sf\ 4x^{2} + 4x - 15 = 0$

Now, Using Factorization method;

$\normalsize\dashrightarrow\sf\ 4x^{2} + 10x - 6x - 15 = 0$

$\normalsize\dashrightarrow\sf\ 2x(2x+5) -3(2x+5) = 0$

$\normalsize\dashrightarrow\sf\ (2x+5)(2x -3)= 0$

$\normalsize\dashrightarrow\sf\ (2x+5) = 0 \: or \: (2x -3) = 0$

$\normalsize\dashrightarrow\sf\ 2x = -5 \: or \: 2x = 3$

$\normalsize\dashrightarrow\sf\ x= \frac{-5}{2} \: or \: \frac{3}{2}$

$\dashrightarrow{\underline{\boxed{\mathsf \pink{x= \frac{-5}{2} \: or \: \frac{3}{2} }}}}$

________________

(b) $\normalsize\sf\sqrt{5}x^{2} + 2x - 3\sqrt{5} = 0$

Using Factorization method;

$\twoheadrightarrow\normalsize\sf\sqrt{5}x^{2} + 5x - 3x - 3\sqrt{5} = 0$

$\twoheadrightarrow\normalsize\sf\sqrt{5}x(x+ \sqrt{5}) - 3( x - \sqrt{5}) = 0$

$\twoheadrightarrow\normalsize\sf\ (x+ \sqrt{5})(\sqrt{5}x - 3)= 0$

$\twoheadrightarrow\normalsize\sf\ (x+ \sqrt{5}) = 0\: or \: (\sqrt{5}x - 3) = 0$

$\twoheadrightarrow\normalsize\sf\ x = 0 - \sqrt{5}\: or \: \sqrt{5}x = 0 + 3$

$\twoheadrightarrow\normalsize\sf\ x = - \sqrt{5}\: or \: \frac{3}{\sqrt{5}}$

$\twoheadrightarrow\normalsize{\underline{\boxed{\mathsf \pink{x = - \sqrt{5}\: or \: \frac{3}{\sqrt{5}} }}}}$

Answer 2

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