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2 years ago

What are the solution of the equation 6x^2-x-1=0

Mathematics

2 answers:

dalvyx [7]2 years ago

4 0

Answer:

Two solutions were found :

6x^2-x1=0

x = -1/3 = -0.333

x = 1/2 = 0.500

Gre4nikov [31]2 years ago

4 0

Answer:

Simplifying

6x2 + x + -1 = 0

Reorder the terms:

-1 + x + 6x2 = 0

Solving

-1 + x + 6x2 = 0

Solving for variable 'x'.

Factor a trinomial.

(-1 + -2x)(1 + -3x) = 0

Step-by-step explanation:

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Find a polynomial of degree 3 with a real coefficient that satisfies the given condition. Zeros: -2, 1, 0; f(2)=32

xz_007 [3.2K]

Its factors would be
(x+2)*(x-1)*(x+0)

x^2 +x -2

x^3 + 0 + x^2 + 0 -2x +0

Equation: x^3 + x^2 -2x
f(2) = 8 + 4 -4

2x^3 + 2x^2 -4x +0
f(2) = 16 + 8 -8

3x^3 + 3x^2 -6x +0
f(2) = 24 +12 -12

4x^3 + 4x^2 -8x +0
f(2) = 32 +16 -16

So, the equation is:
4x^3 + 4x^2 -8x = 0

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2 years ago

What expression is equivalent to 7(7v)

damaskus [11]

Answer:

its the 4th one . The one all the way to the bottom

Step-by-step explanation:

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2 years ago

Read 2 more answers

Write the prime factorization for 65,169

posledela

Let's try to find some primes that divide this number.

The number is not divisible by 2, because it is odd.

The number is divisible by 3 though, because the sum of its digits is:

$6+5+1+6+9=27=3\cdot 9$

So, we can divide the number by 3 and keep going with the factorization:

$65169\div 3 = 21723$

This number is again divisible by 3, because

$2+1+7+2+3 = 15 = 3\cdot 5$

We have

$21723\div 3 = 7241$

This number is no longer divisible by 3. Let's go on looking for primes that divide it: 5 doesn't because the number doesn't end in 0 nor 5. This number is not divisible by 7 or 11 either (just try). It is divisible by 13 though: we have

$7241\div 13 = 557$