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3 years ago

Which two equations would be most appropriately solved by using the quadratic formula?

2 answers:

7 0

Remember that quadratic formula: $x= \frac{-b(+/-) \sqrt{b^{2}-4ac} }{2a}$ is used to solve quadratic equations of the form $ax^{2}+bx+c$ .

Four our problem, the first one, $(x-1)^2=4$ , is not in the form $ax^{2}+bx+c$ . So is not a good idea to use the quadratic formula for this one.

The second one, $0.25x^{2}+0.8x-8=0$ , is in the correct form. Notice that in this equation $a=0.25$ , $b=0.8$ , and $c=-8$ . It would be appropriate to use the quadratic formula to solve this one.

The third one, $3x^{2}-4x=15$ , can be expressed as: $3x^{2}-4x-15=0$ . This equation is also in the correct form. Here $a=3$ , $b=-4$ , and $c=-15$ . It would be appropriate to use the quadratic formula to solve this one.

The fourth one, $(x-6)(x+6)=0$ , is not in the form $ax^{2}+bx+c$ . So is not a good idea to use the quadratic formula for this one.

We can conclude that you should use the quadratic formula to solve the second and third equations.

katrin [286]3 years ago

5 0

Answer: second and third equations, dear.

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1. No
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3 years ago

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The portion of the parabola y²=4ax above the x-axis, where is form 0 to h is revolved about the x-axis. Show that the surface ar

castortr0y [4]

Answer:

See below for Part A.

Part B)

$\displaystyle h=\Big(\frac{125}{\pi}+27\Big)^\frac{2}{3}-9\approx7.4614$

Step-by-step explanation:

Part A)

The parabola given by the equation:

$y^2=4ax$

From 0 to h is revolved about the x-axis.

We can take the principal square root of both sides to acquire our function:

$y=f(x)=\sqrt{4ax}$

Please refer to the attachment below for the sketch.

The area of a surface of revolution is given by:

$\displaystyle S=2\pi\int_{a}^{b}r(x)\sqrt{1+\big[f^\prime(x)]^2} \,dx$

Where r(x) is the distance between f and the axis of revolution.

From the sketch, we can see that the distance between f and the AoR is simply our equation y. Hence:

$r(x)=y(x)=\sqrt{4ax}$

Now, we will need to find f’(x). We know that:

$f(x)=\sqrt{4ax}$

Then by the chain rule, f’(x) is:

$\displaystyle f^\prime(x)=\frac{1}{2\sqrt{4ax}}\cdot4a=\frac{2a}{\sqrt{4ax}}$

For our limits of integration, we are going from 0 to h.

Hence, our integral becomes:

$\displaystyle S=2\pi\int_{0}^{h}(\sqrt{4ax})\sqrt{1+\Big(\frac{2a}{\sqrt{4ax}}\Big)^2}\, dx$

Simplify:

$\displaystyle S=2\pi\int_{0}^{h}\sqrt{4ax}\Big(\sqrt{1+\frac{4a^2}{4ax}}\Big)\,dx$

Combine roots;

$\displaystyle S=2\pi\int_{0}^{h}\sqrt{4ax\Big(1+\frac{4a^2}{4ax}\Big)}\,dx$

Simplify:

$\displaystyle S=2\pi\int_{0}^{h}\sqrt{4ax+4a^2}\, dx$

Integrate. We can consider using u-substitution. We will let:

$u=4ax+4a^2\text{ then } du=4a\, dx$

We also need to change our limits of integration. So:

$u=4a(0)+4a^2=4a^2\text{ and } \\ u=4a(h)+4a^2=4ah+4a^2$

Hence, our new integral is:

$\displaystyle S=2\pi\int_{4a^2}^{4ah+4a^2}\sqrt{u}\, \Big(\frac{1}{4a}\Big)du$

Simplify and integrate:

$\displaystyle S=\frac{\pi}{2a}\Big[\,\frac{2}{3}u^{\frac{3}{2}}\Big|^{4ah+4a^2}_{4a^2}\Big]$

Simplify:

$\displaystyle S=\frac{\pi}{3a}\Big[\, u^\frac{3}{2}\Big|^{4ah+4a^2}_{4a^2}\Big]$

FTC:

$\displaystyle S=\frac{\pi}{3a}\Big[(4ah+4a^2)^\frac{3}{2}-(4a^2)^\frac{3}{2}\Big]$

Simplify each term. For the first term, we have:

$\displaystyle (4ah+4a^2)^\frac{3}{2}$

We can factor out the 4a:

$\displaystyle =(4a)^\frac{3}{2}(h+a)^\frac{3}{2}$

Simplify:

$\displaystyle =8a^\frac{3}{2}(h+a)^\frac{3}{2}$

For the second term, we have:

$\displaystyle (4a^2)^\frac{3}{2}$

Simplify:

$\displaystyle =(2a)^3$

Hence:

$\displaystyle =8a^3$

Thus, our equation becomes:

$\displaystyle S=\frac{\pi}{3a}\Big[8a^\frac{3}{2}(h+a)^\frac{3}{2}-8a^3\Big]$

We can factor out an 8a^(3/2). Hence:

$\displaystyle S=\frac{\pi}{3a}(8a^\frac{3}{2})\Big[(h+a)^\frac{3}{2}-a^\frac{3}{2}\Big]$

Simplify:

$\displaystyle S=\frac{8\pi}{3}\sqrt{a}\Big[(h+a)^\frac{3}{2}-a^\frac{3}{2}\Big]$

Hence, we have verified the surface area generated by the function.

Part B)

We have:

$y^2=36x$

We can rewrite this as:

$y^2=4(9)x$

Hence, a=9.

The surface area is 1000. So, S=1000.

Therefore, with our equation:

$\displaystyle S=\frac{8\pi}{3}\sqrt{a}\Big[(h+a)^\frac{3}{2}-a^\frac{3}{2}\Big]$

We can write:

$\displaystyle 1000=\frac{8\pi}{3}\sqrt{9}\Big[(h+9)^\frac{3}{2}-9^\frac{3}{2}\Big]$

Solve for h. Simplify:

$\displaystyle 1000=8\pi\Big[(h+9)^\frac{3}{2}-27\Big]$

Divide both sides by 8π:

$\displaystyle \frac{125}{\pi}=(h+9)^\frac{3}{2}-27$

Isolate term:

$\displaystyle \frac{125}{\pi}+27=(h+9)^\frac{3}{2}$

Raise both sides to 2/3:

$\displaystyle \Big(\frac{125}{\pi}+27\Big)^\frac{2}{3}=h+9$

Hence, the value of h is:

$\displaystyle h=\Big(\frac{125}{\pi}+27\Big)^\frac{2}{3}-9\approx7.4614$

8 0

3 years ago

Read 2 more answers

Ex 2.11 20) A curve y''=12x-24 and a stationary point at (1,4). evaluate y when x=2.

muminat

So, dy/dx=0 at the point (1, 4) - that is where x=1 and y=4.

$\int { 12x-24dx } \\ \\ =\frac { 12{ x }^{ 2 } }{ 2 } -24x+C\\ \\ =6{ x }^{ 2 }-24x+C$

$\\ \\ \therefore \quad { f }^{ ' }\left( x \right) =6{ x }^{ 2 }-24x+C$

But when x=1, f'(x)=0, therefore:

$0=6-24+C\\ \\ 0=-18+C\\ \\ \therefore \quad C=18$

$\\ \\ \therefore \quad { f }^{ ' }\left( x \right) =6{ x }^{ 2 }-24x+18$

Now:

$\int { 6{ x }^{ 2 } } -24x+18dx\\ \\ =\frac { 6{ x }^{ 3 } }{ 3 } -\frac { 24{ x }^{ 2 } }{ 2 } +18x+C$

$=2{ x }^{ 3 }-12{ x }^{ 2 }+18x+C\\ \\ \therefore \quad f\left( x \right) =2{ x }^{ 3 }-12{ x }^{ 2 }+18x+C$

Now when x=1, y=4:

$4=2-12+18+C\\ \\ 4=8+C\\ \\ C=4-8\\ \\ C=-4$

$\\ \\ \therefore \quad f\left( x \right) =2{ x }^{ 3 }-12{ x }^{ 2 }+18x-4$

Now when x=2,

$f\left( x \right) =2\cdot { 2 }^{ 3 }-12\cdot { 2 }^{ 2 }+18\cdot 2-4\\ \\ =16-48+36-4\\ \\ =0$

So when x=2, y=0.

4 0

3 years ago

Read 2 more answers