How do u estimate ur answer

Not sure what to do here can someone please help

kumpel [21]

Answer:

x = 2

Step-by-step explanation:

These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.

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<h3>Squaring</h3>

The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.

$\sqrt{x+2}+1=\sqrt{3x+3}\qquad\text{given}\\\\(x+2)+2\sqrt{x+2}+1=3x+3\qquad\text{square both sides}\\\\2\sqrt{x+2}=(3x+3)-(x+3)=2x\qquad\text{isolate the root term}\\\\x+2=x^2\qquad\text{divide by 2, square both sides}\\\\x^2-x-2=0\qquad\text{write in standard form}\\\\(x-2)(x+1)=0\qquad\text{factor}$

The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.

x = -1: √(-1+2) +1 = √(3(-1)+3) ⇒ 1+1 = 0 . . . . not true

x = 2: √(2+2) +1 = √(3(2) +3) ⇒ 2 +1 = 3 . . . . true . . . x = 2 is the solution

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<h3>Substitution</h3>

Another way to solve this is using substitution for one of the radicals. We choose ...

$u=\sqrt{x+2}\qquad\text{requires $u\ge0$}\\\\u^2-2=x\qquad\text{solve for x}\\\\u+1=\sqrt{3(u^2-2)+3}\qquad\text{substitute for x in the original equation}\\\\(u+1)^2=3u^2-3\qquad\text{square both sides, simplify a little}\\\\2u^2-2u-4=0\qquad\text{subtract $(u+1)^2$}\\\\2(u-2)(u+1)=0\qquad\text{factor}$

Solutions to this equation are ...

u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution

The value of x is ...

x = u² -2 = 2² -2

x = 2 . . . . the solution to the equation

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<em>Additional comment</em>

Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.

6 0

2 years ago

Which polynomial correctly combines the like terms and puts the given polynomial in standard form? -5x^3y^3+8x^4y^2-xy^5-2x^2y^4

victus00 [196]

Given:

The polynomial is:

$-5x^3y^3+8x^4y^2-xy^5-2x^2y^4+8x^6+3x^2y^4-6xy^5$

To find:

The simplified form of the given polynomial in standard form.

Solution:

We have,

$-5x^3y^3+8x^4y^2-xy^5-2x^2y^4+8x^6+3x^2y^4-6xy^5$

Combining the like terms, we get

$=-5x^3y^3+8x^4y^2+(-2x^2y^4+3x^2y^4)+8x^6+(-xy^5-6xy^5)$

$=-5x^3y^3+8x^4y^2+x^2y^4+8x^6-7xy^5$

Now, rewrite this polynomial in standard form.

$=8x^6+8x^4y^2-5x^3y^3+x^2y^4-7xy^5$

Therefore, the required polynomial in standard form is $8x^6+8x^4y^2-5x^3y^3+x^2y^4-7xy^5$ .

8 0

3 years ago

If the width of something is 5m and the perimeter is 22m, what is the length and area

Nadusha1986 [10]

I assume that this 'something' has a rectangular shape. If its width is 5m, than the length of two sides is 2 * 5 = 10m. Now if the perimeter is 22 square meters, then the lenght of another two sides is 12 / 2 = 6.

12 because 22 (the perimeter) - 10 (lenght of the 2 sides) = 12 (length of another 2 sides).

4 0

3 years ago

Solve for all possible values of x. Show all work.<br> X + 2<br> 2x - 1<br> 3x

pogonyaev

Answer:

a) x+2=0

x= -2

b) 2x-1=0

2x=1

x= 1/2

c) 3x=0

x= 0/3

x=0

6 0

3 years ago

PLEASE SOMEONE ASAP HELP ME!!!!!!!!!!!

MrRissso [65]

She distributed incorrectly.

7 0

3 years ago