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3 years ago

Factor to find the zeros of the function defined by the quadratic expression. 9x2 − 63x − 702

Mathematics

2 answers:

labwork [276]3 years ago

6 0

Answer:

The zeros of the quadratic expression are: x=-6 and x=13

Step-by-step explanation:

First we need to use the common factor:

$9x^2-63x-702=9(x^2-7x-78)$

Now factorizing the expression of the parentheses we have:

$9(x^2-7x-78)=9(x+6)(x-13)$

To know the zeros we need to equal to zero the expression then:

$9(x+6)(x-13)=0$

$(x+6)(x-13)=0$

This expression will be zero when x=-6 or when x=13.

yan [13]3 years ago

5 0

Not that it will help all that much, but you could take out 9 as a common factor.
9(x^2 - 7x - 78) = 0 so since 9 but itself can't produce a zero, we can divide by 9
x^2 - 7x - 78 = 0
There are 3 factors of 78
2 13 and 3
6 and 13 differ by 7 (6 comes from putting the 2 and 3 together by multiplying.
(x - 13)(x + 6) = 0

x -13 =0
x = 13

x + 6 = 0
x = - 6

The two zeros are (-6,0) and (13,0)

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3 years ago

Solve the system y = -x + 7 and y = 0.5(x - 3)2

amid [387]

Answer:

(5, 2 ) and (- 1, 8 )

Step-by-step explanation:

Given the equations

y = - x + 7 → (1)

y = 0.5(x - 3)²

= 0.5(x² - 6x + 9)

y = 0.5x² - 3x + 4.5 → (2)

Substitute y = - x + 7 into (2)

- x + 7 = 0.5x² - 3x + 4.5 ( subtract - x + 7 from both sides )

0 = 0.5x² - 2x - 2.5 ( multiply through by 2 ) , then

x² - 4x - 5 = 0 ← in standard form

(x - 5)(x + 1) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 5 = 0 ⇒ x = 5

x + 1 = 0 ⇒ x = - 1

Substitute these values into (1) for corresponding values of y

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3 years ago

<img src="https://tex.z-dn.net/?f=%20%20%5Cdisplaystyle%20%5Cint%20%5Climits_%7B0%7D%5E%7B%20%5Cfrac%7B%20%5Cpi%7D%7B2%7D%20%7D%

murzikaleks [220]

Let $x = \arcsin(y)$ , so that

$\sin(x) = y$

$\tan(x)=\dfrac y{\sqrt{1-y^2}}$

$dx = \dfrac{dy}{\sqrt{1-y^2}}$

Then the integral transforms to

$\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_{y=\sin(0)}^{y=\sin\left(\frac\pi2\right)} \frac{y}{\sqrt{1-y^2}} \ln(y) \frac{dy}{\sqrt{1-y^2}}$

$\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy$

Integrate by parts, taking

$u = \ln(y) \implies du = \dfrac{dy}y$

$dv = \dfrac{y}{1-y^2} \, dy \implies v = -\dfrac12 \ln|1-y^2|$

For 0 < y < 1, we have |1 - y²| = 1 - y², so

$\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = uv \bigg|_{y\to0^+}^{y\to1^-} + \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy$

It's easy to show that uv approaches 0 as y approaches either 0 or 1, so we just have

$\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy$

Recall the Taylor series for ln(1 + y),

$\displaystyle \ln(1+y) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n y^n$

Replacing y with -y² gives the Taylor series

$\displaystyle \ln(1-y^2) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n (-y^2)^n = - \sum_{n=1}^\infty \frac1n y^{2n}$

and replacing ln(1 - y²) in the integral with its series representation gives

$\displaystyle -\frac12 \int_0^1 \frac1y \sum_{n=1}^\infty \frac{y^{2n}}n \, dy = -\frac12 \int_0^1 \sum_{n=1}^\infty \frac{y^{2n-1}}n \, dy$

Interchanging the integral and sum (see Fubini's theorem) gives

$\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy$

Compute the integral:

$\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy = -\frac12 \sum_{n=1}^\infty \frac{y^{2n}}{2n^2} \bigg|_0^1 = -\frac14 \sum_{n=1}^\infty \frac1{n^2}$

and we recognize the famous sum (see Basel's problem),

$\displaystyle \sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}6$

So, the value of our integral is

$\displaystyle \int_0^{\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \boxed{-\frac{\pi^2}{24}}$

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