Question

2x² + 5x +2 = 0 solve by
Factorization method.​

Answer 1

Given : 5x² - 3x - 2 = 0

5x² - 3x - 2 = 0

5x² - 5x + 2x - 2 = 0

5x(x - 1) +2 (x - 1) = 0

(5x + 2)(x - 1) = 0

5x + 2 = 0 or x - 1= 0

5x = - 2 or x = 1

x = -2/5 or x = 1✌✌

Hence, the roots of the quadratic equation 5x² - 3x - 2 = 0 are - ⅖ & 1 .

★★ METHOD TO FIND SOLUTION OF a quadratic equation by FACTORIZATION

METHOD :

We will first write the given quadratic polynomial as product of two linear factors by splitting the middle term and then equate each factor to zero to get desired roots of given quadratic equation.

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Answer 2

Answer:

Required value of x is $( - 2) \: or \: ( - \frac{1}{2} )$

Step-by-step explanation:

Given,

$2{x}^{2} + 5x + 2 = 0$

By factorization method,

$2 {x}^{2} + (4 + 1)x + 2 = 0 \\ 2 {x}^{2} + 4x + x + 2 = 0 \\ 2x(x + 2) + (x + 2) = 0 \\ (x + 2)(2x + 1) = 0$

We know product of two terms is zero then they are separately zero.

So,

$x + 2 = 0 \\ x = - 2$

and

$2x + 1 = 0 \\ 2x = - 1 \\ x = \frac{ - 1}{2} \\ x = - \frac{1}{2}$

So, required value of x is $( - 2) \: or \: ( - \frac{1}{2} )$

This is a problem of Algebra.

Some important Algebra formulas:

${(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \\ {(x + y)}^{3} = {x}^{3} + 3 {x}^{2} y + 3x {y}^{2} + {y}^{3} \\ {(x - y)}^{3} = {x}^{3} - 3 {x}^{2} y + 3x {y}^{2} - {y}^{3} \\ {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y) \\ {x}^{3} - {y}^{3} = {(x - y)}^{3} + 3xy(x - y) \\ {x}^{2} - {y}^{2} = (x + y)(x - y) \\ {x}^{2} + {y}^{2} = {(x - y)}^{2} + 2xy \\ {x}^{2} - {y}^{2} = {(x + y)}^{2} - 2xy \\ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} ) \\ {x}^{3} + {y}^{3} = (x - + y)( {x}^{2} - xy + {y}^{2} )$

Know more about Algebra,

1) https://brainly.in/question/13024124

2) https://brainly.in/question/1169549

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