Question

2x square-7x +3=0(solve by complete squaring method)​

Answer 1

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$\large\purple{\sf 2x^{2} - 7x + 3 = 0}$

Dividing by 2

$\longrightarrow$ $\large\sf \frac{2x^{2} \:- \:7x\: + \:3 \:=\: 0}{2} = \frac{0}{2}$

$\longrightarrow$ $\large\sf \frac{\cancel{2x}^{2}}{\cancel 2} - \frac{7x}{2} + \frac{3}{2} = 0$

$\longrightarrow$ $\large\sf x^{2} - \frac{7x}{2} + \frac{3}{2} = 0$

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We know that,

$\large\orange{\sf (a - b)^{2} = a^{2} - 2ab + b^{2}}$

Here, a = x and

$\longrightarrow$ $\large\sf - 2ab = \frac{7x}{2}$

$\longrightarrow$ $\large\sf - 2xb = \frac{7x}{2}$ ( As a = x )

$\longrightarrow$ $\large\sf b = - \frac{7x}{2(-2x)}$

$\longrightarrow$ $\large\sf b = \frac{7}{4}$

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Now, in equation

$\longrightarrow$ $\large\sf x^{2} - \frac{7x}{2} + \frac{3}{2}$

Adding and substracting $\large\sf (\frac{7}{4})^{2}$

$\longrightarrow$ $\large\sf x^{2} - \frac{7x}{2} + \frac{3}{2} + (\frac{7}{4})^{2} - (\frac{7}{4})^{2}= 0$

$\longrightarrow$ $\large\sf x^{2} + (\frac{7}{4})^{2} - \frac{7x}{2} + \frac{3}{2} - (\frac{7}{4})^{2} = 0$

$\longrightarrow$ $\large\sf \left( x - \frac{7}{4} \right)^{2} + \frac{3}{2} - (\frac{7}{4})^{2} = 0$

$\longrightarrow$ $\large\sf \left( x - \frac{7}{4} \right)^{2} + \frac{3}{2} - \frac{49}{16} = 0$

$\longrightarrow$ $\large\sf \left( x - \frac{7}{4} \right)^{2} + \frac{3(8) - 49}{16} = 0$

$\longrightarrow$ $\large\sf \left( x - \frac{7}{4} \right)^{2} + \frac{24 - 49}{16} = 0$

$\longrightarrow$ $\large\sf \left( x - \frac{7}{4} \right)^{2} - \frac{25}{16} = 0$

$\longrightarrow$ $\large\sf \left( x - \frac{7}{4} \right)^{2} = \frac{25}{16}$

$\longrightarrow$ $\large\sf \left( x - \frac{7}{4} \right)^{2} = (\frac{5}{4})^{2}$

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Cancelling square both sides :-

$\mapsto$ $\large\sf x - \frac{7}{4} = \pm \frac{5}{4}$

Case 1 :-

$\blue\leadsto$ $\large\blue{\sf x - \frac{7}{4} = \frac{5}{4}}$

$\blue\leadsto$ $\large\blue{\sf x = \frac{5}{4} + \frac{7}{4}}$

$\blue\leadsto$ $\large\blue{\sf x = \frac{5 + 7}{4}}$

$\blue\leadsto$ $\large\blue{\sf x = \frac{12}{4}}$

$\pink\leadsto$ $\large\boxed{\pink{\sf x = 3}}$ $\orange\star$

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Case 2 :-

$\green\leadsto$ $\large\green{\sf x -\frac{7}{4} = -\frac{5}{4}}$

$\green\leadsto$ $\large\green{\sf x = - \frac{5}{4} + \frac{7}{4}}$

$\green\leadsto$ $\large\green{\sf x = \frac{-5 + 7}{4}}$

$\green\leadsto$ $\large\green{\sf x = \frac{2}{4}}$

$\pink\leadsto$ $\large\boxed{\pink{\sf \frac{1}{2}}}$ $\orange\star$

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So the roots of the quadratic equation are

x = 3 and x = 1/2.

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