Question

find the 4x sq +3x+5=0 by completing the square mathod​

Answer 1

$\sf{\underline{\boxed{\big{\pink{Question}}}}}$

• Solve by completing the square $4x^2 + 3x + 5 = 0$

$\sf{\underline{\boxed{\big{\pink{Solution}}}}}$

$\sf\implies 4x^2 +3x + 5 = 0$

$\sf\implies 4x^2 + 3x = -5$

• Divide each term by 4

$\sf\implies \dfrac{4x^2}{4} + \dfrac{3x}{4} = \dfrac{-5}{4}$

$\sf\implies x^2 + \dfrac{3x}{4}= \dfrac{-5}{4}$

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

$\sf\longrightarrow(\dfrac{3}{4\times 2})^2$

$\sf\longrightarrow(\dfrac{3}{8})^2$

• add the term to each side of the equation

$\sf\implies x^2 + \dfrac{3x}{4} + (\dfrac{3}{8})^2 = \dfrac{-5}{4} + ( \dfrac{3}{8})^2$

$\sf\implies x^2 + \dfrac{3x}{4} + \dfrac({3}{8})^2 = \dfrac{-5}{4} + \dfrac{ 9 }{64}$

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

$\sf\implies ( x + \dfrac{3}{8})^2 = \dfrac{-5 \times 16 + 9}{64}$

$\sf\implies ( x + \dfrac{5}{6})^2 = \dfrac{-80+ 9}{64}$

$\sf\implies ( x + \dfrac{5}{6})^2 = \dfrac{-71}{64}$

• since the roots of any number cannot be negative
• Therefore this eq. have no roots

$\sf{\underline{\boxed{\big{\purple{Equation\: have\: no \: roots }}}}}$