Question

5x^2+3x-2=0 solved by factor and completing the square​

Answer 1

Solution :-

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

$\sf\rightarrow 5{x}^{2} + (5 - 2)x - 2 = 0$

$\sf\rightarrow 5{x}^{2} + 5x - 2x - 2 = 0$

$\sf\rightarrow 5x(x + 1) - 2(x + 1) = 0$

$\sf\rightarrow (5x - 2)(x + 1) = 0$

$\sf\therefore Either, (5x - 2) = 0 \: or, (x + 1) = 0$

$\sf Solving \: for \: the \: value \: of \: x,$

$\sf\rightarrow 5x - 2 = 0$

$\sf\rightarrow 5x = 2$

$\sf\therefore x = \dfrac{2}{5}$

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$\sf\rightarrow x + 1 = 0$

$\sf\therefore x = -1$

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$\sf\therefore Required \: value \: of \: x = \: \dfrac{2}{5} \: or, \: -1$

Now, Verifying if the values of x satisfies the equation,

$\sf\rightarrow 5 \times {\dfrac({2}{5})}^{2} + 3 \times \dfrac{2}{5} - 2 = 0$

$\sf\rightarrow 5 \times \dfrac{4}{25} + \dfrac{6}{5} - 2 = 0$

$\sf\rightarrow \dfrac{20}{25} + \dfrac{6}{5} - 2 = 0$

$\sf\rightarrow \dfrac{4}{5} + \dfrac{6}{5} - 2 = 0$

$\sf\rightarrow \dfrac{4 + 6 - 10}{5} = 0$

$\sf\rightarrow \dfrac{10 - 10}{5} = 0$

$\sf\rightarrow \dfrac{0}{5} = 0$

$\sf\rightarrow 0 = 0$

Putting the second value,

$\sf\rightarrow 5 \times {-1}^{2} + 3 \times (-1) - 2 = 0$

$\sf\rightarrow 5 \times 1 + (-3) - 2 = 0$

$\sf\rightarrow 5 - 3 - 2 = 0$

$\sf\rightarrow 5 - 5 = 0$

$\sf\rightarrow 0 = 0$

$\sf\therefore Verified$

Answer 2

Answer:

Factorization method:

5x²+3x-2=0

5x²+5x-2x-2=0

5x(x+1) -2(x+1) =0

(5x-2) (x+1) =0

therefore, x=2/5 or -1