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avanturin [10]
3 years ago
6

Factor the quadratic equation below to reveal the solutions. X^2+4x-21=-9

Mathematics
1 answer:
a_sh-v [17]3 years ago
6 0

Answer:

x = 3 and x = -7

Step-by-step explanation:

The given quadratic equation is x^2+4x-21=0. We need to find the solution of this equation.

If the equation is in the form of ax^2+bx+c=0, then its solutions are given by :

x=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}

Here, a = 1, b = 4 and c = -21

Plugging all the values in the value of x, such that :

x=\dfrac{-b+ \sqrt{b^2-4ac} }{2a},\dfrac{-b- \sqrt{b^2-4ac} }{2a}\\\\x=\dfrac{-4+ \sqrt{(4)^2-4\times 1\times (-21)} }{2(1)},\dfrac{-4- \sqrt{(4)^2-4\times 1\times (-21)} }{2(1)}\\\\x=3, -7

So, the solutions of the quadratic equation are 3 and -7.      

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Answer:

2 x 3 + 6 - 8

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2 x 3 = 6

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12 - 8 = 4

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The management of an electronics firm has determined that the profit the company earns is a function of thenumber of employees i
ruslelena [56]

we have the function

P(x)=\ln\left|4x+1\right|+3x-x^2

For x=2

substitute

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10 months ago
Hello i need help with this question
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4pf

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Which of the shaded regions below represents 3/5 of the whole circle
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Answer: B

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2 years ago
The sum of first three terms of a finite geometric series is -7/10 and their product is -1/125. [Hint: Use a/r, a, and ar to rep
Alchen [17]
Ooh, fun

geometric sequences can be represented as
a_n=a(r)^{n-1}
so the first 3 terms are
a_1=a
a_2=a(r)
a_2=a(r)^2

the sum is -7/10
\frac{-7}{10}=a+ar+ar^2
and their product is -1/125
\frac{-1}{125}=(a)(ar)(ar^2)=a^3r^3=(ar)^3

from the 2nd equation we can take the cube root of both sides to get
\frac{-1}{5}=ar
note that a=ar/r and ar²=(ar)r
so now rewrite 1st equation as
\frac{-7}{10}=\frac{ar}{r}+ar+(ar)r
subsituting -1/5 for ar
\frac{-7}{10}=\frac{\frac{-1}{5}}{r}+\frac{-1}{5}+(\frac{-1}{5})r
which simplifies to
\frac{-7}{10}=\frac{-1}{5r}+\frac{-1}{5}+\frac{-r}{5}
multiply both sides by 10r
-7r=-2-2r-2r²
add (2r²+2r+2) to both sides
2r²-5r+2=0
solve using quadratic formula
for ax^2+bx+c=0
x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}
so
for 2r²-5r+2=0
a=2
b=-5
c=2

r=\frac{-(-5) \pm \sqrt{(-5)^2-4(2)(2)}}{2(2)}
r=\frac{5 \pm \sqrt{25-16}}{4}
r=\frac{5 \pm \sqrt{9}}{4}
r=\frac{5 \pm 3}{4}
so
r=\frac{5+3}{4}=\frac{8}{4}=2 or r=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}

use them to solve for the value of a
\frac{-1}{5}=ar
\frac{-1}{5r}=a
try for r=2 and 1/2
a=\frac{-1}{10} or a=\frac{-2}{5}


test each
for a=-1/10 and r=2
a+ar+ar²=\frac{-1}{10}+\frac{-2}{10}+\frac{-4}{10}=\frac{-7}{10}
it works

for a=-2/5 and r=1/2
a+ar+ar²=\frac{-2}{5}+\frac{-1}{5}+\frac{-1}{10}=\frac{-7}{10}
it works


both have the same terms but one is simplified

the 3 numbers are \frac{-2}{5}, \frac{-1}{5}, and \frac{-1}{10}
6 0
3 years ago
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