Question

Find the roots of 4x2 + 3x + 5 = 0 by the method of completing the square

Answer 1

4x2 + 3x + 5 =0
or, ${x} = (- (-3) + \sqrt{ (-3)^{2} - 4.4.5}) / 2.4 or (- (-3) - \sqrt{ (-3)^{2} - 4.4.5}) / 2.4$
or, x = ${x} = (3 + \sqrt{ (-71}) / 8 or (3 - \sqrt{ (-71}) / 8$

Answer 2

$Given:-$

• $equation:- {{4x}^{2} + {3x} + {5}}$

$To \: \: Find:-$

• $Root \: \: of \: \: the \: \: given \: \: equation \: \: by \: \: completing \: \: the \: \: square \: \: method.$

$Solutions:-$

• ${{4x}^{2} + {3x} + {5}}$

$⇢ {{4x}^{2} + {3x} + {5}}$

• $Divible \: \: both \: \: side \: \: by \: \: 4.$

$⇢ \sf{x}^{2} + \frac{3x}{4} + \frac{5}{4} = \frac{0}{4}$

$⇢ {x}^{2} + \frac{3x}{4}+ \frac{5}{4} = {0}$

• $Add \: \: {(\frac{3}{8})}^{2} \: \: on \: \: both \: \: side.$

$⇢ \sf{x}^{2} + \frac{3x}{4} + \frac{9}{64} = \frac{-5}{4} + \frac{9}{64}$

$⇢ {({x} + \frac{3x}{8})}^{2} = \frac{-71}{64}$

$⇢ \sf{x} + \frac{3x}{8} = \frac{\sqrt{-71}}{64}$

$⇢ \sf{x} + \frac{3x}{8} = \frac{{+-}{\sqrt{71}}}{8}$

$⇢ {x} = \frac{{+-}{\sqrt{71}}}{8} - \frac{3x}{8}$

$x = \frac{\sqrt{71} - {3}}{8} \: \: \: or \: \: \: x = \frac{\sqrt{-71} + {3}}{8}$

$So, \: \: the \: \: root \: \: of \: \: the \: \: given \: \: equation \: \: are:-$

• $x = \frac{\sqrt{71} - {3}}{8}$

• $x = \frac{\sqrt{-71} + {3}}{8}$

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