All categories

3 years ago

2x^2+9x+6=0 quadtratic equation

Mathematics

1 answer:

Studentka2010 [4]3 years ago

5 0

Hello from MrBillDoesMath!

Answer:

x = ( -9 + sqrt(33) ) /4

and

x = ( -9 - sqrt(33) ) /4

Discussion:

Using the quadratic formula with a = 2, b = 9 and c = 6 gives

x = ( -b +\- sqrt(b^2-4ac) )/2a

x = ( -9 +\- sqrt(9^2 - 4*2*6)) / (2*2) =>

x = ( -9 +\- sqrt (81- 48)) / 4 =>

x = ( -9 +\- sqrt(33) ) /4

Thank you,

MrB

You might be interested in

Anne and Kyleigh worked over summer vacation. The equation below represents the amount of money Anne earned after x hours of wor

Dovator [93]

Kyleigh earns more per hour

3 0

3 years ago

If a and b are real numbers, what property of real numbers is illustrated by the equation below? 4a + 4b = 4(a + b)

Tresset [83]

D. Distributed Property of Multiplication over Addition. Happy Thanksgiving. Hope this helps.

7 0

3 years ago

Can someone please answer these questions???!

Leto [7]

Answer:

Step-by-step explanation:

sub in all your x values into the linear equation it gives you

once you have your table you should be able to figure out the rest if the questions !!

7 0

3 years ago

50 squared equals 2 + 25 squared equals 2 * 5 equals 10 what is wrong

Klio2033 [76]

Convert to an equation:

50^2=2+25^2=2*5=10

Simplify:

2500=625=10=10

What is wrong is that 2500 does not equal 625 which does not equal 10.

7 0

3 years ago

Find the smallest positive $n$ such that \begin{align*} n &\equiv 3 \pmod{4}, \\ n &\equiv 2 \pmod{5}, \\ n &\equiv

Alex777 [14]

4, 5, and 7 are mutually coprime, so you can use the Chinese remainder theorem right away.

We construct a number $x$ such that taking it mod 4, 5, and 7 leaves the desired remainders:

$x=3\cdot5\cdot7+4\cdot2\cdot7+4\cdot5\cdot6$

Taken mod 4, the last two terms vanish and we have

$x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod4$

so we multiply the first term by 3.

Taken mod 5, the first and last terms vanish and we have

$x\equiv4\cdot2\cdot7\equiv51\equiv1\pmod5$

so we multiply the second term by 2.

Taken mod 7, the first two terms vanish and we have

$x\equiv4\cdot5\cdot6\equiv120\equiv1\pmod7$

so we multiply the last term by 7.

Now,

$x=3^2\cdot5\cdot7+4\cdot2^2\cdot7+4\cdot5\cdot6^2=1147$

By the CRT, the system of congruences has a general solution

$n\equiv1147\pmod{4\cdot5\cdot7}\implies\boxed{n\equiv27\pmod{140}}$

or all integers $27+140k$ , $k\in\mathbb Z$ , the least (and positive) of which is 27.

3 0

3 years ago