How can I solve this equation 5x^2-2x-6=-2x^2+9x

1 answer:

wel2 years ago

3 0

Add 2x^2 to both sides
7x^2-2x-6=9x

Subtract 9x from both sides
7x^2-11x-6=0

Factor by using slip and slide
x^2-11x-42
(x-14)(x+3)
(x-14/7)(x+3/7)
(x-2)(7x+3)

Then use the Zero Product Property to find roots
x-2=0
x=2

7x+3=0
7x=-3
x=-3/7

Final answer: x=-3/7, x=2

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What is the sum of the infinite geometric series 9−6+4−83+...9-6+4-83+...??

vivado [14]

Successive terms have a common factor of $-\dfrac23$ , which means the $n$ th partial sum of the series can be written as

$S_n=9-6+4-\dfrac83+\cdots+9\left(-\dfrac23\right)^{n-1}$
$S_n=9\left(1+\left(-\dfrac23\right)^1+\left(-\dfrac23\right)^2+\left(-\dfrac23\right)^3+\cdots+\left(-\dfrac23\right)^{n-1}$

$\implies-\dfrac23S_n=9\left(\left(-\dfrac23\right)^1+\left(-\dfrac23\right)^2+\left(-\dfrac23\right)^3+\left(-\dfrac23\right)^4+\cdots+\left(-\dfrac23\right)^n\right)$

$\implies S_n-\left(-\dfrac23\right)S_n=9\left(1-\left(-\dfrac23\right)^n\right)$
$\dfrac53S_n=9\left(1-\left(-\dfrac23\right)^n\right)$
$S_n=\dfrac{27}5\left(1-\left(-\dfrac23\right)^n\right)$

As $n\to\infty$ , the geometric term vanishes, leaving you with

$S=\displaystyle\lim_{n\to\infty}S_n=\dfrac{27}5$