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Zepler [3.9K]
3 years ago
14

Solve the following equations by factorisation method.Only factorisation not dharacharya.​

Mathematics
1 answer:
Luba_88 [7]3 years ago
6 0

Hello, please consider the following.

When x_1 and x_2 are two roots, we can factorise as

ax^2+bx+c=a(x-x_1)(x-x_2)=a(x^2-(x_1+x_2)x+x_1x_2)=0

So for the first equation, we can say that the sum of the zeros is

\dfrac{a^2+b^2}{2}=\dfrac{a^2}{2}+\dfrac{b^2}{2}

and the product is

\dfrac{a^2b^2}{4}=\dfrac{a^2}{2}\dfrac{b^2}{2}

So we can factorise as below.

4x^2-2(a^2+b^2)x+a^2b^2=(2x-a^2)(2x-b^2)=0

And the solutions are

\boxed{\sf \n\bf  \ \dfrac{a^2}{2} \ \ and \ \ \dfrac{b^2}{2}}

For the second equation, we will complete the square and put the constant on the right side and take the root.

Let's do it!

9x^2-9(a+b)x+2a^2+5ab+2b^2=0\\\\9(x-\dfrac{a+b}{2})^2-9\dfrac{(a+b)^2}{4}+2a^2+5ab+2b^2=0\\\\9(x-\dfrac{a+b}{2})^2=\dfrac{9(a+b)^2-8a^2-20ab-8b^2}{4}\\\\9(x-\dfrac{a+b}{2})^2=\dfrac{9a^2+18ab+9b^2-8a^2-20ab-8b^2}{4}\\\\9(x-\dfrac{a+b}{2})^2=\dfrac{a^2+b^2-2ab}{4}=\dfrac{(a-b)^2}{4}\\\\(x-\dfrac{a+b}{2})^2=\dfrac{(a-b)^2}{2^2\cdot 3^2}

We take the root, and we find the two solutions

\begin{aligned}x_1&=\dfrac{a+b}{2}-\dfrac{a-b}{6}\\\\&=\dfrac{3a+3b-a+b}{6}\\\\&=\dfrac{2a+4b}{6}\\\\&\boxed{=\dfrac{a+2b}{3}}\end{aligned}

\begin{aligned}x_1&=\dfrac{a+b}{2}+\dfrac{a-b}{6}\\\\&=\dfrac{3a+3b+a-b}{6}\\\\&=\dfrac{4a+2b}{6}\\\\&\boxed{=\dfrac{2a+b}{3}}\end{aligned}

Thank you.

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An object is thrown straight up from the top of a 100-foot platform at a velocity of 48 feet per second. The height h(t) of the
julsineya [31]

Answer:

  • The maximum height is 136 ft
  • The time it takes to achieve this height is 1.5 s.

Explanation:

<u>1. Function for the height (given):</u>

    h(t)=-16t^2+48t+100

<u />

<u>2. Type of function</u>

That is a quadatic function, whose graph is a parabola that opens downward.

The maximum of the function, i.e. the maximum height, is the vertex of the parabola.

The vertex of a parabola with the genral equation   y=ax^2+bx+c  is at the x-coordinate

                       x=-b/(2a)

<u>3. Time to achieve the maximum height</u>

Substitute b with 48 and a with - 16:

        t=-48/(2(-16))=48/32=3/2=1.5

Then, time when the object achieves the maximum height it 1.5s

<u />

<u>4. Maximum height:</u>

Replace t with 1.5 in the equation, to find the maximum height, h(1.5)

     h(1.5)=-16(1.5)^2+48(1.5)+100=136

Then, the maximum height is 136 ft

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The radius of the base of the cylinder is 2x cm and the height of the cylinder is h cm.
valentina_108 [34]

Answer:

h = 9x

Step-by-step explanation:

Given: i. for the cylinder, base radius = 2x cm, height = h cm

           ii. for the sphere, radius = 3x cm

           iii.  volume of the cylinder = volume of the sphere

volume of a cylinder is given as;

volume = \frac{1}{3}\pir^{2}h

where: r is its base radius and h the height

volume of the given cylinder = \frac{1}{3}\pi x (2x)^{2} x h

                                              = \frac{1}{3}\pi x 4x^{2} x h

                                              = \frac{4}{3}\pix^{2} h

volume of a sphere = \frac{4}{3}\pir^{3}

where r is the radius.

volume of the given sphere =  \frac{4}{3}\pi x (3x)^{3}

                                         =  \frac{4}{3}\pi x 9 x^{3}

                                         = 12\pix^{3}

Since,

volume of the cylinder = volume of the sphere

Then we have;

\frac{4}{3}\pix^{2} h = 12\pix^{3}

4\pix^{2} h = 36\pix^{3}

subtract x^{2} from both sides

4\pih = 36\pix

divide both sides by 4\pi

h = \frac{36\pi x}{4\pi }

  = 9x

h = 9x

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Answer:

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Step-by-step explanation:

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