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# 5under root xsquare +2x+3` ` ` ` ` ` `

user6426479 | eNotes Newbie

Posted November 16, 2013 at 7:36 AM via web

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5under root xsquare +2x+3`

` ` ` ` ` `

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted November 16, 2013 at 9:03 AM (Answer #2)

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You need to reduce the given expression to its lowest terms, hence, since the expression contains `x^2` under radical, hence, you need to use the following definition, such that:

`sqrt(x^2) = |x|`

Using the definition of absolute value, yields:

`|x| = +-x`

Hence, you will have to answers after the evaluation of expression, such that:

Considering `|x| = x` , yields:

`5sqrt(x^2) + 2x + 3 = 5x + 2x + 3 `

Adding the coefficients of terms that contain x yields:

`5sqrt(x^2) + 2x + 3 = 7x + 3`

Considering `|x| = -x` , yields:

`5sqrt(x^2) + 2x + 3 = -5x + 2x + 3 `

`5sqrt(x^2) + 2x + 3 =-3x + 3`

Hence, evaluating the expression, you will have two answers, such that: `5sqrt(x^2) + 2x + 3 = 7x + 3` for `x > 0` and `5sqrt(x^2) + 2x + 3 =-3x + 3` , for `x < 0` .

aruv | High School Teacher | Valedictorian

Posted November 19, 2013 at 7:03 PM (Answer #3)

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Square root is function. Thus `sqrt(x^2)!=|x|` .

So

`sqrt(x^2)!=|x|` `={(x if x>=0),(-x if x<0):}`

Otherwise we can see it from following graph.

Let define a function `f(x)=sqrt(x^2)` , the domain of f(x) is all real numbers and co domain is only positive real numbers. Its graph is given below

Let define a function `f(x)=-sqrt(x^2)` , the domain of f(x) is all real numbers and co domain is only negative real numbers. Its graph is given below

Thus from above graphs we can conclude that

`5sqrt(x^2)+2x+3=5x+2x+3`

`=7x+3`

It is true for all values of `x inRR` and if `5(-sqrt(x^2))+2x+3` then

`5(-sqrt(x^2))+2x+3=-5x+2x+3`

`=-3x+3`

` AA x in RR` .

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