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dolphi86 [110]
3 years ago
5

What is the number of real solutions? –11x^2 = x + 11

Mathematics
2 answers:
Anvisha [2.4K]3 years ago
8 0
-11 x² = x + 11
11 x² + x + 11 = 0
We will find the value of the discriminant:
D = b² - 4 a c = 1² - 4 * 11 * 11 * 1 - 484 = - 483
D < 0. There are no real solutions.
Aleks04 [339]3 years ago
8 0

Answer:

Given the equation: -11x^2 = x+11

or we can write above equation as:

-11x^2-x-11=0 or

11x^2+x+11 =0

The general equation of quadratic formula for ax^2+bx+c=0; where a, b and c are constant;

Use Discriminant formula: D =b^2-4ac

If D> 0, then there are 2 roots.

If D = 0, then there is only 1 root.

If D <0, then there are no real roots.

Now, from the given equation 11x^2+x+11 =0 we have

a =11, b= 1 and c =11

Then, using discriminant formula we have;

D =b^2-4ac =(1)^2 -4(11)(11) = 1-4\cdot 121 = 1-484= -483

⇒D<0 [ No real roots]

Therefore, for the given equation -11x^2 = x+11 , there are no real solutions.


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What are the x-intercepts of the parabola?
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(5,0) & (4,0)

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What are the next 3 numbers in this pattern of integers -2, -1, -1, 0, -1, 1, -2
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Answer:

The next 3 numbers to the given pattern of integers is -2,-1,-1,0,-1,1,-2 are 3,-5,8

Step-by-step explanation:

Given pattern of integers is -2,-1,-1,0,-1,1,-2

To find their next 3 numbers of the given pattern :

Let a_{1}=-2

a_{2}=-1

a_{3}=a_{1}-a_{2}

=-2-(-1)

=-2+1

Therefore a_{3}=-1

a_{4}=a_{2}-a_{3}

=-1-(-1)

=-1+1

Therefore a_{4}=0

Similarly we have a_{5}=-1 ,a_{6}=1 ,a_{7}=-2

The general form to the given pattern is

a_{n}=a_{n-2}+a_{n-1}  for n={\{1,2,3,4,5,6,7,8,9,10}\}

Therefore to find the a_{8},a_{9},a_{10}

Put n=8 in a_{n}=a_{n-2}-a_{n-1}

a_{8}=a_{8-2}-a_{8-1}

=a_{6}-a_{7}

=1-(-2)

a_{8}=3

Put n=9 in a_{n}=a_{n-2}-a_{n-1}

a_{9}=a_{9-2}-a_{9-1}

=a_{7}-a_{8}

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a_{9}=-5

Put n=10 in a_{n}=a_{n-2}-a_{n-1}

a_{10}=a_{10-2}-a_{10-1}

=a_{8}-a_{9}

=3-(-5)

a_{10}=8

Therefore the next 3 numbers are 3,-5,8

4 0
3 years ago
81/372 in simplest form
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