Question

Solve by completing the square 3x2 -- 5x + 2 = 0

Answer 1

$\sf{\large{\underline{\boxed{\pink{\mathsf{♡ Question ♡}}}}}}$

• Solve by completing the square $3x^2 + 5x - 2 = 0$

$\sf{\large{\underline{\boxed{\pink{\mathsf{♡ solution ♡}}}}}}$

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

$\sf\implies 3x^2 + 5x = 2$

• Divide each term by 3

$\sf\implies \dfrac{3x^2}{3} + \dfrac{5x}{3} = \dfrac{2}{3}$

$\sf\implies x^2 + \dfrac{5x}{3}= \dfrac{2}{3}$

• Divide the coefficient of x by 2, and square it and add it on both sides in the equation

$\sf\longrightarrow(\dfrac{5}{3\times 2})^2$

$\sf\longrightarrow(\dfrac{5}{6})^2$

• add the term to each side of the equation

$\sf\implies x^2 + \dfrac{5x}{3} + (\dfrac{5}{6})^2 = \dfrac{2}{3} + ( \dfrac{ 5}{6})^2$

$\sf\implies x^2 + \dfrac{5x}{3} + \dfrac( {5}{6} )^2 = \dfrac{2}{3} + \dfrac{ 25}{36}$

$\sf{\large{\underline{\boxed{\red{\mathsf{a^2+ 2ab + b^2 = ( a + b )^2 }}}}}}$

$\sf\implies ( x + \dfrac{5}{6})^2 = \dfrac{2 \times 12 + 25}{36}$

$\sf\implies ( x + \dfrac{5}{6})^2 = \dfrac{24 + 25}{36}$

$\sf\implies ( x + \dfrac{5}{6})^2 = \dfrac{49}{36}$

$\sf\implies ( x + \dfrac{5}{6}) = \sqrt{\dfrac{49}{36}}$

$\sf\implies ( x + \dfrac{5}{6})= \pm\dfrac{7}{6}$

• solve for x

$\sf\implies x + \dfrac{5}{6} = \dfrac{7}{6}$

$\sf\implies x = \dfrac{7}{6} - \dfrac{5}{6}$

$\sf\implies x = \dfrac{7 - 5}{6}$

$\sf\implies x = \dfrac{2}{6}$

$\sf\implies x = \dfrac{1}{3}$

$\sf{\large{\underline{\boxed{\green{\mathfrak{x = \dfrac{1}{3}}}}}}}$

$\sf\implies x + \dfrac{5}{6} = - \dfrac{7}{6}$

$\sf\implies x = - \dfrac{7}{6} - \dfrac{5}{6}$

$\sf\implies x = \dfrac{- 7 - 5}{6}$

$\sf\implies x = \dfrac{- 12 }{6}$

$\sf\implies x = - 2$

$\sf{\large{\underline{\boxed{\green{\mathfrak{x = - 2 }}}}}}$

$\sf{\large{\underline{\boxed{\orange{\mathfrak{x = \dfrac{1}{3} \: and - 2 }}}}}}$

Answer 2

${\huge{\boxed{\red{\mathcal{Hello}}}}}$

Solve by completing the square 3x2 -- 5x + 2 = 0

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