Question

7x²+3x-4=0 solve by completing square method

Answer 1

$\underline {\pink { Completing \:square \:method :}}$

$Given \: Quadratic\: Equation: \\7x^{2} + 3x - 4 = 0$

/* Dividing each term by 7, we get */

$\implies x^{2} + \frac{3}{7} x - \frac{4}{7} = 0$

$\implies x^{2} + \frac{3}{7} x = \frac{4}{7}$

$\implies x^{2} +2 \times x \times \frac{3}{2\times 7} = \frac{4}{7}$

$\implies x^{2} +2 \times x \times \frac{3}{14} = \frac{4}{7}$

$Add\: both\: sides\: by \: \Big(\frac{3}{14}\Big)^{2},\:we \:get$

$\implies x^{2} +2 \times x \times \frac{3}{14}+ \Big(\frac{3}{14}\Big)^{2} = \Big(\frac{3}{14}\Big)^{2} + \frac{4}{7}$

$\implies \Big( x + \frac{3}{14}\Big)^{2} = \frac{9}{196} + \frac{4}{7}$

$\implies \Big( x + \frac{3}{14}\Big)^{2} = \frac{9}{196} + \frac{4}{7}$

$\implies \Big( x + \frac{3}{14}\Big)^{2} = \frac{252+4}{196}$

$\implies \Big( x + \frac{3}{14}\Big)^{2} = \frac{256}{196}$

$\implies \Big( x + \frac{3}{14}\Big)^{2} = \frac{16^{2}}{14^{2}}$

$\implies \Big( x + \frac{3}{14}\Big)^{2} =\Big( \frac{16}{14}\Big)^{2}$

$\implies x + \frac{3}{14}=\pm \frac{16}{14}$

$\implies x = -\frac{3}{14}\pm \frac{16}{14}$

$\implies x = \frac{-3\pm16}{14}$

$\implies x = \frac{-3+16}{14} \: Or \: x = \frac{-3- 16}{14}$

$\implies x = \frac{13}{14} \: Or \: x = \frac{-19}{14}$

Therefore.,

$\green { x = \frac{13}{14} \: Or \: x = \frac{-19}{14}}$

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Answer 2

Answer:

-1 , 4/7

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