genyjoannerubiera

01.02.2021 •

Mathematics

The first term of the sequence is -3 and the common ratio
is -1/4

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Ответ:

shaheedbrown06

12.01.2022

Mathematics

$m=\cfrac{\sqrt{73}\pm 7 }{4}$

Step-by-step explanation:

$2m^2\:-\:7m\:-\:13\:=\:-10$

We'll use Quadratic formula to solve this equation:

$\cfrac{-b\pm \sqrt{b^2-4ac} }{2a}$

______________________

2m^2-7m-13=-10

→ Add 10 from both sides:

-13-(-10)=-10-(-10)

→ Subtract -10 from -10= 0

2m^2-7m-13-(10)=0

→ Subtract -10 from -13:

Now, Equation is in standard from ax^2+bx+c=0, we'll solve it using Quadratic formula:

$\cfrac{-b\pm \sqrt{b^2-4ac} }{2a}$

$m=\cfrac{-\left(-7\right)\pm \sqrt{\left(-7\right)^2-4\times \:2\left(-3\right)}}{2\times \:2}$

Square -7, you'll get 49, multiply -4 * 2= 8, and than -8 * -3= 24

$m=\cfrac{-(-7)\pm \sqrt{49+24} }{2\times 2}$

→ Add 49 + 24= 73

$m=\cfrac{-(-7)\pm \sqrt{73} }{2\times 2}$

$m=\cfrac{7\pm\sqrt{73} }{2\times 2}$

→ Multiply 2 * 2= 4

$m=\cfrac{\sqrt{73}\pm 7 }{4}$

__________________________

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The first term of the sequence is -3 and the common ratio is -1/4