Solve the inequality 2x^2 + 4x >= 6

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embizze | High School Teacher | (Level 1) Educator Emeritus

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Solve `2x^2+4x>=6 ` :

`2x^2+4x>=6 `

`2x^2+4x-6>=0 `

`2(x^2+2x-3)>=0 ` Since 2>0 we look at the quadratic:

`(x+3)(x-1)>=0 `

For x<-3, the left side is positive.

For -3<x<1 the left side is negative.

For x>1 the left side is positive.

(In each case, try a test value such as x=-4,x=0,x=2)

Thus the solution intervals are `x<=-3 ` or `x>=1 `

In interval notation the solution is `(-oo,-3]uu[1,oo) `

The graph:

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The inequality `2x^2 + 4x >= 6` has to be solved.

`2x^2 + 4x >= 6`

An inequality is not affected if the same constant is added or subtracted from both the sides.

`2x^2 + 4x - 6 >= 6 - 6`

=> `2x^2 + 4x - 6 >= 0`

=> `2x^2 + 6x - 2x - 6 >= 0`

=> `2x(x - 3) - 2(x - 3) >= 0`

=> `(x - 1)(x - 3) >= 0`

The product (x - 1)(x - 3) is greater than or equal to 0 if either both the terms are positive or both the terms are negative.

`x - 1 >= 0` , `x - 3>= 0`

=> `x >= 3`

x - 1 < 0, x - 3 < 0

=> x < 1

The inequality is satisfied for all values of x lying in the set `(-oo, 1]U(3, oo)`

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justaguide | College Teacher | (Level 2) Distinguished Educator

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Sorry the errors in the response. Please refer the correct response posted by embizze.

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