Question

Solve the quadratic equation 9x2-9x+2=0 completing square method

Answer 1

Answer:

⅓ and ⅔

Step-by-step explanation:

Given a quadratic equation such that,

$9 {x}^{2} - 9x + 2 = 0$

To solve this fir x.

We need to follow completing square method.

Therefore, we will get,

$= > {(3x)}^{2} - 2 (3x)( \frac{3}{2} ) + {( \frac{3}{2} )}^{2} + 2 - {( \frac{3}{2}) }^{2} = 0$

But, we know that,

• a^2 -2ab + b^2 = (a-b)^2

Therefore, we will get,

$= > {(3x - \frac{3}{2} )}^{2} + 2 - \frac{9}{4} = 0 \\ \\ = > {(3x - \frac{3}{2}) }^{2} + \frac{8 - 9}{4} = 0 \\ \\ = > {(3x - \frac{3}{2}) }^{2} - \frac{1}{4} = 0 \\ \\ = > {(3x - \frac{3}{2}) }^{2} = \frac{1}{4} \\ \\ = > 3x - \frac{3}{2} = \pm \frac{1}{2} \\ \\ = > 3x = \frac{3}{2} \pm \frac{1}{2} \\ \\ = > 3x = 2 \: \: \: and \: \: \: 1 \\ \\ = > x = \frac{1}{3} \: \: and \: \: \frac{2}{3}$

Hence, the value of x is ⅔ and ⅓.

Answer 2

Given:-

${9x} ^{2} - 9x+2=0$

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By method of Completing square:-

${(3x)}^{2} - 2 (3x)( \frac{3}{2} ) + {( \frac{3}{2} )}^{2} + 2 - {( \frac{3}{2}) }^{2} = 0$

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$⇝ {(3x - \frac{3}{2} )}^{2} + 2 - \frac{9}{4} = 0 \\ ⇝ {(3x - \frac{3}{2}) }^{2} + \frac{8 - 9}{4} = 0 \\ ⇝ {(3x - \frac{3}{2}) }^{2} - \frac{1}{4} = 0 \\ ⇝ {(3x - \frac{3}{2}) }^{2} = \frac{1}{4} \\ ⇝3x - \frac{3}{2} = \pm \frac{1}{2} \\ ⇝3x = \frac{3}{2} \pm \frac{1}{2} \\⇝ 3x = 2 \: and \: 1 \\ ⇝ x = \frac{1}{3} \: and \: \frac{2}{3}$

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∴ $\red{x= \frac{2}{3}~and~\frac{1}{3}}$.