Question

only correct answers ..plz​

Answer 1

Answer:

x = 0 it should be correct

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Answer 2

$\sf\huge\bold{Answer:}$

Given :

$\tt\ \frac{x + 2}{x - 3} - \frac{x + 7}{x + 2} = \frac{25}{6}$

To find :

Simplifying to quadratic equation in standard form

Solution :

$\tt\ :⟶ \frac{x + 2}{x - 3} - \frac{x + 7}{x + 2} = \frac{25}{6}$

• Taking LCM

$\tt\ :⟶ \frac{x + 2( x + 2) - (x + 7)(x - 3)}{(x - 3)(x + 2)} = \frac{25}{6}$

$\tt\ \: :⟶ \frac{(x + 2) {}^{2} - x(x - 3) - 7(x - 3) }{x(x + 2) - 3(x + 2)} = \frac{25}{6}$

$\bf\red {(a + b) {}^{2} = {a}^{2} + {b}^{2} + 2ab }$

$\tt\ :⟶ \frac{ {x}^{2} + {2}^{2} + (2 \times 2 \times x )- {x}^{2} + 3x - 7x + 21}{ {x}^{2} + 2x - 3x - 6 } = \frac{25}{6}$

$\tt\ :⟶ \frac{ {x}^{2} + 4 + 4x - {x}^{2} + 3x - 7x + 21 }{ {x}^{2} - x - 6 } = \frac{25}{6}$

$\tt\ :⟶ \frac{ \cancel{{x}^{2}} - \cancel{{x}^{2}} + 4x + 3x - 7x + 4 + 21}{ {x}^{2} - x - 6 } = \frac{25}{6}$

$\tt\ :⟶\frac{\cancel{4x} - \cancel{4x} + 25}{ {x}^{2} - x - 6} = \frac{25}{6}$

$\tt\ :⟶ \frac{25}{ {x}^{2} - x - 6 } = \frac{25}{6}$

$\tt\ :⟶ \frac{6}{ {x}^{2} - x - 6} = \frac{25}{25}$

$\tt{:⟶ \frac{6}{x^2-x-6}=}$$\displaystyle{\sf { \cancel{ \frac{25}{25} }}}$

$\tt\ :⟶ \frac{6}{ {x}^{2} - x - 6} = 1$

$\tt\ :⟶6 = 1( {x}^{2} - x - 6)$

$\tt\ :⟶6 = {x}^{2} - x - 6$

$\tt\ :⟶ {x}^{2} - x - 6 = 6$

$\tt\ :⟶ {x}^{2} - x - 6 - 6 = 0$

$\tt\ :⟶ {x}^{2} - x - 12 = 0$

• Standard form of quadratic equation → ax²+bx+c=0

Quadratic equation in standard form is

$\huge \boxed{ \boxed{\tt\purple{{x}^{2} - x - 12} = 0}}$