Determine whether x=-1/2 x=1/3 are the solutions of the given equation 6x2-x-2=0 or not

Step-by-step explanation : Check whether the following are quadratic equations : (i) (x + 1)2 = 2(x – 3)

In each of the following determine whether the given values are solutions of the equation or not.
6x2 - x - 2 = 0; x = `-(1)/(2), x = (2)/(3)`

Given equation
6x2 - x - 2 = 0; x = `-(1)/(2), x = (2)/(3)`Substitute x = `-(1)/(2)` in L.H.S.L.H.S. = `6(-1/2)^2 - (-1/2)-2`= `6 xx (1)/(4) + (1)/(4) - 2`= 2 - 2= 0Hence, x = `-(1)/(2)` is a solution of the given equation.Also put x = `(2)/(3)` in L.H.S.L.H.S. = `6(2/3)^2 - (2)/(3) -2`= `6 xx (4)/(9) - (2)/(3) - 2`= `(8)/(3) - (2)/(3) -2`= `(6)/(3) - 2` = 2 - 2= 0

Hence, x = `(2)/(3)` is a solution of the given equation.

Concept: Solutions of Quadratic Equations by Factorization

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Page 2

In each of the following determine whether the given values are solutions of the equation or not.
x2 + `sqrt(2)` - 4 = 0; x = `sqrt(2)`, x = -2`sqrt(2)`

Given equation
x2 + `sqrt(2)` - 4 = 0; x = `sqrt(2)`, x = -2`sqrt(2)`Substitute x = `sqrt(2)` in the L.H.S.L.H.S. = `(sqrt(2))^2 + sqrt(2) xx sqrt(2) - 4`= 2 + 2 - 4= 4 - 4= 0

Hence x = `sqrt(2)` is a solution of the given equation.

Concept: Solutions of Quadratic Equations by Factorization

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Page 3

In each of the following determine whether the given values are solutions of the equation or not.
9x2 - 3x - 2 = 0; x = `-(1)/(3), x = (2)/(3)`

Given equation is 
9x2 - 3x - 2 = 0; x = `-(1)/(3), x = (2)/(3)`Substitute x = `-(1)/(3)` in the L.H.S.L.H.S. = `9(-1/3)^2 - 3 xx (-1/3) -2`= `9 xx (1)/(9) + 1 - 2`= 2 - 2= 0= R.H.S.Hence, x = `-(1)/(3)` is a solution of the equation.Again put x = `(2)/(3)`L.H.S. = `9(2/3)^2 -3(2/3)-2`= `9 xx (4)/(9) - 2 -2`= 4 - 4= 0= R.H.S.

Hence, x = `(2)/(3)` is a solution of the equation.

Concept: Solutions of Quadratic Equations by Factorization

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Page 4

In each of the following determine whether the given values are solutions of the equation or not.
x2 + x + 1 = 0; x = 1, x = -1.

Given equation is
x2 + x + 1 = 0; x = 1, x = -1Substitute x = 1 in L.H.S.

L.H.S. = (1)2 + (1) + 1

L.H.S. = (-1)2 + (-1) + 1

Hence, x = -1 is not a solution of the given equation.

Concept: Solutions of Quadratic Equations by Factorization

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Page 5

In each of the following, determine whether the given values are solution of the given equation or not:
x2 - 3x + 2 = 0; x = 2, x = -1

Substitute x = 2 in L.H.S. of given equation
L.H.S. = (2)2 - 3x 2 x2= 6 - 6= 0⇒ L.H.S. = 0= R.H.S.Substitute x = -1 in L.H.S. of given equation.

L.H.S. = (-1)2 - 3 x -1 + 2 = 0

x = 2 is a solution and x = -1 is not solution of the given equation.

Concept: Solutions of Quadratic Equations by Factorization

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Page 6

In each of the following, determine whether the given values are solution of the given equation or not:
x2 + x + 1 = 0; x = 0; x = 1

Now Substitute x = 0 in given equation
L.H.S. = (0)2 + 0 + 1 ≠ 0 ≠ R.H.S.on substituting x = 1 in L.H.S. of given equation

⇒ (1)2 + 1 + 1 ≠ 0 ≠ R.H.S.

Concept: Solutions of Quadratic Equations by Factorization

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Page 7

In each of the following, determine whether the given values are solution of the given equation or not:
x2 - 3`sqrt(3)` + 6 = 0; x = `sqrt(3)`, x = -2`sqrt(3)`

x2 - 3`sqrt(3)` + 6; x = `sqrt(3)`, x = -2`sqrt(3)`.Now substitute x = `sqrt(3)` in L.H.S. of given equation.L.H.S. = `(sqrt(3))^2 -3sqrt(3) xx sqrt(3) + 6 = 0`= 3 - 9 + 6= 0= R.H.S.x = `sqrt(3)` is a solution of the given equation.Substitute x = `-2sqrt(3)` in L.H.S. of given equation⇒ `(-2sqrt(3))^2 -3sqrt(3) xx -2sqrt(3) + 6 = 0`⇒ L.H.S. = 12 + 18 + 6 ≠ 0 ≠ R.H.S.

x = `-2sqrt(3)` is not a solution of the given equation.

Concept: Solutions of Quadratic Equations by Factorization

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