Question

12. Solve the quadratic equation 3x2 + 7x + 1 = 0 by
the method of completing the square

​

Answer 1

Solution :-

→ 3x² + 7x + 1 = 0

Step 1 :- Take the constant term alone on the right side.

→ 3x² + 7x = (-1)

step 2 :- Divide both sides by the coefficient of x² term.

→ (3x²/3) + (7x/3) = (-1/3)

→ x² + (7x/3) = (-1/3)

step 3 :- Add the square of half the coefficient of the x term, to both sides .

• Half of coefficient of the x term = Half of (7/3) = (1/2) * (7/3) = (7/6) .

→ x² + (7x/3) + (7/6)² = (-1/3) + (7/6)²

step 4 :- Factor the left side as the square of a binomial , and solve right side.

→ (x)² + 2 * x * (7/6) + (7/6)² = (49/36) - (1/3)

→ (x)² + 2 * x * (7/6) + (7/6)² = (49 - 12)/36

→ (x)² + 2 * x * (7/6) + (7/6)² = (37/36)

comparing LHS with a² + 2ab + b² = (a + b)² , we get,

→ (x + 7/6)² = (37/36)

Step 5 :- Square - root both sides Now,

→ (x - 7/6) = ± (√37/6)

→ x = ± (√37/6) + (7/6)

→ x = (7 ± √37)/6

→ x = (7 + √37)/6 or , (7 - √37)/6. (Ans.)

Answer 2

Step-by-step explanation:

Given:

$3 {x}^{2} + 7x + 1 = 0$

To find:Solve the quadratic equation by

the method of completing the square

Solution:

To find the solution by completing square method,

1. put constant term to other hand
2. add a term both sides which convert LHS in complete square

$3 {x}^{2} + 7x = - 1 \\ \\ ( { \sqrt{3}x) }^{2} + 2( \sqrt{3}x)\bigg( \frac{7}{2 \sqrt{3} } \bigg)+\bigg(\frac{7}{2 \sqrt{3} }\bigg)^{2} = - 1+\bigg(\frac{7}{2 \sqrt{3} }\bigg)^{2} \\ \\ {\bigg( \sqrt{3}x + \frac{7}{2 \sqrt{3} } \bigg) }^{2} = - 1 + \frac{49}{12} \\ \\ {\bigg( \sqrt{3}x + \frac{7}{2 \sqrt{3} } \bigg) }^{2} = \frac{ - 12 + 49}{12} \\ \\{\bigg( \sqrt{3}x + \frac{7}{2 \sqrt{3} } \bigg) }^{2} = \frac{ 37}{12} \\ \\ {\bigg( \sqrt{3}x + \frac{7}{2 \sqrt{3} } \bigg) }^{2} = \bigg(\sqrt{\frac{ 37}{12}}\bigg)^{2}$

Apply identity a²-b²

${\bigg( \sqrt{3}x + \frac{7}{2 \sqrt{3} } \bigg) }^{2} -\bigg(\sqrt{\frac{ 37}{12}}\bigg)^{2} = 0 \\ \\ \bigg( \sqrt{3} x + \frac{7}{2 \sqrt{3}} + \frac{\sqrt{37} }{2 \sqrt{3} }\bigg)\bigg( \sqrt{3} x + \frac{7}{2 \sqrt{3}} - \frac{\sqrt{37} }{2 \sqrt{3} }\bigg) = 0 \\ \\ \bigg( \sqrt{3} x + \frac{7}{2 \sqrt{3}} + \frac{\sqrt{37} }{2 \sqrt{3} }\bigg) = 0 \\ \\ x = \frac{ - 7 - \sqrt{37} }{2 \sqrt{3} \times \sqrt{3} } \\ \\ \bold{x = \frac{ - 7 - \sqrt{37} }{6} } \\ \\ or \\ \\ \bigg( \sqrt{3} x + \frac{7}{2 \sqrt{3}} - \frac{\sqrt{37} }{2 \sqrt{3} }\bigg) = 0 \\ \\ \bold{x = \frac{ \sqrt{37} - 7}{6}}$

Hope it helps you.

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