Question

Find the roots of the quadratic equation 6x2

+5x + 1 =0 by method of completing

the squares.​

Answer 1

Given,

The given quadratic equation = 6x²+5x+1 = 0

To find,

Solving by the method of completing the squares.

Solution,

First of all, we have to keep only the "x" related terms on the LHS of the equation.

So,

6x²+5x+1 = 0

6x²+5x = -1

Now, we have to convert the coefficient of "x²" into 1.

So,

6x²/6 + 5x/6 = -1/6

x²+5x/6 = -1/6....(1)

Now,we have to take the coefficient of x and then we have to divide that coefficient by 2. Finally we need to square the result of the previous division.

Coefficient of x = 5/6

Half of the coefficient = 5/12

Square of the division = (5/12)²

Now, we need to add the result of the "Square of the division", on the both sides of equation (1).

x²+5x/6+ (5/12)² = -1/6 + 25/144

(x + 5/12)² = 25/144 -1/6

(x + 5/12)² = (25-24)/144

(x + 5/12)² = 1/144

(x + 5/12)² = (1/12)²

x + 5/12 = 1/12

x = 1/12-5/12

x = (1-5)/12

x = -4/12

x = -1/3

Hence, the value of x will be -1/3

Answer 2

SOLUTION

TO DETERMINE

The roots of the quadratic equation

$\sf{6 {x}^{2} + 5x + 1 = 0 }$

by method of completing the squares

EVALUATION

Here the given Quadratic equation is

$\sf{6 {x}^{2} + 5x + 1 = 0 }$

Since it is a quadratic equation

So it has two roots

$\sf{6 {x}^{2} + 5x + 1 = 0 }$

Dividing both sides by 6 we get

$\displaystyle \: \sf{ {x}^{2} + \frac{5x}{6} + \frac{1}{6} = 0 }$

$\implies \displaystyle \: \sf{ {x}^{2} + 2.x. \frac{5}{12} + { \bigg( \frac{5}{12} \bigg)}^{2} - { \bigg( \frac{5}{12} \bigg)}^{2} + \frac{1}{6} = 0 }$

$\implies \displaystyle \sf{ { \bigg( x + \frac{5}{12} \bigg)}^{2} - \frac{25}{144} + \frac{1}{6} = 0 }$

$\implies \displaystyle \sf{ { \bigg( x + \frac{5}{12} \bigg)}^{2} - \frac{1}{144} = 0 }$

$\implies \displaystyle \sf{ { \bigg( x + \frac{5}{12} \bigg)}^{2} = \frac{1}{144} }$

$\implies \displaystyle \sf{ { \bigg( x + \frac{5}{12} \bigg)} = \pm \: \frac{1}{12} }$

Now

$\displaystyle \sf{ { \bigg( x + \frac{5}{12} \bigg)} = \frac{1}{12} } \: \: \: gives$

$\displaystyle \sf{x = \frac{1}{12} - \frac{5}{12} }$

$\implies\displaystyle \sf{x = - \frac{4}{12} }$

$\implies\displaystyle \sf{x = - \frac{1}{3} }$

Again

$\displaystyle \sf{ { \bigg( x + \frac{5}{12} \bigg)} = - \frac{1}{12} } \: \: \: gives$

$\displaystyle \sf{x = - \frac{1}{12} - \frac{5}{12} }$

$\implies\displaystyle \sf{x = - \frac{6}{12} }$

$\implies\displaystyle \sf{x = - \frac{1}{2} }$

Hence the required roots of the given quadratic equation are

$\displaystyle \sf{ - \frac{1}{3} \: \: \: and \: \: - \frac{1}{2} }$

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