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3 years ago

Can someone help me solve (x+18)^(1/2)+2=x

1 answer:

6 0

Lets solve your equation step by step:
(x+18)^1/2+2=x
step 1:Add -2 to both sides:
(x+18)^1/2+2-2=x+-2
(x+18^)1/2=x-2
step 2:solve exponent
(x+18)^1/2=x-2
((x+18)^1/2)^2=(x-2)^2
x+18=x^2-4x+4
x+18-(x^2-4x+4)=x^2-4x+4-(x^2-4x+4) (subtract x^2-4x+4 from both sides)
-x^2+5x+14=0
(-x-2)(x-7)=0(factor left side of equation)
-x-2=0 or x-7=0 (set factors equal to zero)
x=-2 or x=7

check answers:(plug them in to make sure they work )
x=-2(doesn't work in original equation)
x=7(works in original equation)

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Why are halves a good choice for benchmark fractions for 1 1/3

Natalka [10]

Answer and explanation:

Benchmark fractions are fractions that are used as references in measuring other fractions. They are easily estimated and so can be used in measuring more "specific" fractions such as 1/5, 7/9, 3/7, 1/3 etc. If I wanted to measure 1 1/3cm for instance using a calibrated ruler, having centimeter measurements, I would first find 1cm on the ruler and then find half of one centimeter. Seeing that half is bigger than 1/3 but close, I could then estimate 1/3 to be somewhere less than 1/2 but a bit close to it

3 0

3 years ago

What are the solutions of the equation (2x + 3)2 + 8(2x + 3) + 11 = 0

Tom [10]

Answer:

$x=\frac{-7-\sqrt{5}}{2}$ or $x=\frac{-7+ \sqrt{5}}{2}$

Step-by-step explanation:

The given equation is

$(2x+3)^2+8(2x+3)+11=0$

Let us treat this as a quadratic equation in $(2x+3)$ .

where $a=1,b=8,c=11$

The solution is given by the quadratic formula;

$(2x+3)=\frac{-b\pm \sqrt{b^2-4ac} }{2a}$

We substitute these values into the formula to obtain;

$(2x+3)=\frac{-8\pm \sqrt{8^2-4(1)(11)} }{2(1)}$

$(2x+3)=\frac{-8\pm \sqrt{64-44} }{2}$

$(2x+3)=\frac{-8\pm \sqrt{20} }{2}$

$(2x+3)=\frac{-8\pm2\sqrt{5} }{2}$

$(2x+3)=-4\pm \sqrt{5}$

$(2x+3)=-4-\sqrt{5}$ or $(2x+3)=-4+ \sqrt{5}$

$2x=-3-4-\sqrt{5}$ or $2x=-3-4+ \sqrt{5}$

$2x=-7-\sqrt{5}$ or $2x=-7+ \sqrt{5}$

$x=\frac{-7-\sqrt{5}}{2}$ or $x=\frac{-7+ \sqrt{5}}{2}$

5 0

3 years ago

Find x. Assume that segments that appear tangent are tangent.

kipiarov [429]

Assuming that CB is tangent, this is just a right triangle with a hypotenuse of 20 and a side of 12 so by the Pythagorean Theorem:

20^2=12^2+x^2

400=144+x^2

x^2=400-144

x^2=256

x=16

6 0

3 years ago

What is 4 radical 3 divided by 2?

Arte-miy333 [17]

3/2 = 6
6/4= 6/40= 6 .999= ~ 7

5 0

3 years ago

A container is shaped like a triangle prism. Each base of container is an equilateral triangle with the dimensions shown. The co

Aleks04 [339]

Answer:

$\text{Lateral surface area of container}=270\text{ cm}^2$

Step-by-step explanation:

Please find the attachment.

We have been given that a container is shaped like a triangle prism. Each base of container is an equilateral triangle with each side 6 cm. The height of container is 15 cm.

To find the lateral surface area of our given container we will use lateral surface area formula of triangular prism.

$\text{Lateral surface area of triangular prism}=(a+b+c)*h$ , where, a, b and c represent base sides of prism and h represents height of the prism.

Upon substituting our given values in above formula we will get,

$\text{Lateral surface area of container}=(\text{6 cm+ 6 cm+6 cm})*\text{15 cm}$

$\text{Lateral surface area of container}=\text{18 cm}*\text{15 cm}$

$\text{Lateral surface area of container}=270\text{ cm}^2$

Therefore, lateral surface area of our given container is 270 square cm.

6 0

3 years ago