Math, asked by madhurigole, 10 months ago

solve the following equation in complete the square method m square - 14 + 13 is equal to zero​

Answers

Answered by Sharad001
50

Question :-

Solve the following quadratic equation by completing the square method

→ m² - 14m + 13 = 0

Answer :-

→ m = 13 and 1

Explanation :-

We have quadratic equation

→ m² - 14m + 13 = 0

Completing the square method .

(•) Step (1) -

Firstly divide the whole equation by the coefficient of m² ( already done )

→ Make 1 as coefficient of m² .

(•)Step (2) -

Keep the constant term in right side

→ m² - 14m = -13

(•) Step (3) -

Complete the square of left side of the equation and balance this equation by adding and subtracting the half of coefficient of m i.e 14/2 = 7

 \to \sf{  {m}^{2}  - 14m +  {(7)}^{2} -  {(7)}^{2}   =  - 13} \\  \\  \to \sf{ {m}^{2}  +  {(7)}^{2}  - 2 \times 7m =  - 13 + 49} \\   \\  \because \bf{ {a}^{2}  +   \red{{b}^{2}  }- 2ab =  \green{ {(a - b)}^{2} }} \\  \\  \to \sf{  {(m - 7)}^{2}  = 36} \\   \\  \to \sf{{(m - 7)}^{2}  =  { ( \pm6)}^{2} } \\  \\  \bf{ \underline{case \: (1)}}  \:  \: if \:  +  \: \\  \\  \implies \sf{m - 7 = 6} \\  \\  \implies \boxed{ \sf{m = 13}} \\  \\  \bf{ \underline{case \: (2)} \: if \:  -  \: } \\  \\  \implies \sf{m - 7 =  - 6} \\  \\  \implies \:   \boxed{\sf{m = 1}}

hence ,m = 1 and 13 .

Verification :-

If m = 1

→ 1 -14 + 13 = 0

→ 0 = 0

if m = 13

→ 169 - 14×13 + 13 = 0

→ 169 - 172 +13 = 0

→ 0= 0

Hence verified.

Answered by Saby123
2

 \tt{\huge{\orange {Hello!!! }}}

QUESTION :

Solve the following equation in complete the square method m square - 14 m + 13 is equal to zero.

SOLUTION :

The above Answer shows the perfect solution using the " Completing The Square Method "

So I am not going to repeat the same boring solution and solve in a much easier and shorter way by factorising...

F { X } = { M } ^ 2 - 14 { M } + 13.

=> M ^ 2 - 13 M - M + 13

=> M ( M - 13 ) - 1 ( M - 13 )

=> ( M - 1 ) ( M - 13 )

Hence M = 1 and M = 13 are the solutions of the equation.

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