Question

find the roots of x square - 7x + 12 = 0 by factorization method ​

Answer 1

Answer:

x²-7x+12=0

x²-4x-3x+12=0. (By splitting the middle term)

x(x-4)-3(x-4)=0

(x-3)(x-4)=0

When,

x-3=0

x=3

When,

x-4=0

x=4

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Answer 2

Given : Quadratic equation = x² - 7x + 12

To Find : The roots of the Given Quadratic equation

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⠀⠀⠀⠀⠀$\sf{\bf{ Solution \:of\:Question \::}}$

⠀⠀⠀⠀⠀$\longmapsto {\mathrm {Equation \: = \:x^{2} - 7x + 12 = 0}}$

By using Sum - Product Pattern :

⠀⠀⠀⠀⠀⠀⠀⠀$\longmapsto {\mathrm { \:x^{2} \purple {- 7x} + 12 = 0}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\longmapsto {\mathrm { \:x^{2}\purple{- 4x - 3x } + 12 = 0}}$

⠀⠀⠀⠀⠀⠀⠀⠀$\longmapsto {\mathrm { \:x^{2}- 4x - 3x + 12 = 0}}$

Now , By Finding factor in Each term :

⠀⠀⠀⠀⠀⠀⠀⠀$\longmapsto {\mathrm { \:x^{2}- 4x - 3x + 12 = 0}}$

Now , Taking x as Common in First term :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\longmapsto {\mathrm { \:\purple {x^{2}- 4x } - 3x + 12 = 0}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\longmapsto {\mathrm { \:\purple {x( x - 4) } - 3x + 12 = 0}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\longmapsto {\mathrm { \:x( x - 4) - 3x + 12 = 0}}$

Now , Taking -3 as Common in Second term :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\longmapsto {\mathrm { \:x( x - 4) \purple { - 3x + 12} = 0}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\longmapsto {\mathrm { \:x( x - 4) \purple { - 3( x - 4 ) } = 0}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀$\longmapsto {\mathrm { \:x( x - 4) - 3( x - 4 ) = 0}}$

Now , Rewrite in Factored term :

⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀$\longmapsto {\mathrm { \purple {\:x( x - 4) - 3( x - 4 )} = 0}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀$\longmapsto {\mathrm {\purple {\:( x - 3) ( x - 4 ) }= 0}}$

As , We know that ,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\longmapsto {\mathrm { x - 3= 0}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\underline {\boxed {\pink{\mathrm { x = 3 }}}}$

and ,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\longmapsto {\mathrm { x - 4 = 0}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\underline {\boxed {\pink{\mathrm { x = 4}}}}$

$\dag\:\:{\underline {\pink{\mathrm { Hence ,\: The \:roots \:of\:x^{2} - 7x + 12 = 0 \:are \:3 \:and\: 4.}}}}$

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