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Bingel [31]
3 years ago
9

Which are the solutions of x^2 = –11x + 4?

Mathematics
1 answer:
strojnjashka [21]3 years ago
3 0

This is a quadratic equation, i.e. an equation involving a polynomial of degree 2. To solve them, you must rearrange them first, so that all terms are on the same side, so we get

x^2 + 11x - 4 = 0

i.e. now we're looking for the roots of the polynomial. To find them, we can use the following formula:

x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2}

where x_{1,2} is a compact way to indicate both solutions x_1 and x_2, while a,b,c are the coefficients of the quadratic equation, i.e. we consider the polynomial ax^2+bx+c.

So, in your case, we have a=1,\ \ b=11,\ \ c=-4

Plug those values into the formula to get

x_{1,2} = \frac{-11\pm\sqrt{121+16}}{2} = \frac{-11\pm\sqrt{137}}{2}

So, the two solutions are

x_1 = \frac{-11+\sqrt{137}}{2}

x_2 = \frac{-11-\sqrt{137}}{2}

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irga5000 [103]

Answer:

a_{n} = 8 - 11n

Step-by-step explanation:

These are the terms of an arithmetic sequence with n th term

a_{n} = a + (n - 1)d

where a is the first term and d the common difference

here a = - 3 and

d = - 14 - (- 3) = - 14 + 3 = -11, thus

a_{n} = - 3 - 11(n - 1)

                        = - 3 - 11n + 11

                        = 8 - 11n

3 0
3 years ago
Find the equation of a straight line passing throught the point listed and having the given gradient. Express your answer in the
Igoryamba
First let's try to find the equation in this form : <span>y = mx + c 

The gradient is given 3 . In a line's equation, x's coefficient represents the line's gradient.

So equation of a line with the gradient of 3, would look like this ;

</span>y= 3x + c
<span>
Now a point that the line passes through is given, (1, 2) 

This point's x-coordinate is 1 and y-coordinate is 2.

So we'll plug its x-coordinate value in the equation and also y-coordinate value. So we can solve it.

As you know, </span>x=1 and y=2

y = 3x + c

2\quad =\quad 3\cdot 1+c\\ \\ 2\quad =\quad 3+c\\ \\ 2-3\quad =\quad c\\ \\ -1\quad =\quad c

We found c = -1 

Also in a line's equation, c is constant and it represents the line's y-intercept

So let's build the line's equation.

m=3 and c=-1

y= mx + c

y= 3x -1

We found the line's equation in this form, y= mx + c

Now let's turn it into this form, ax + by + c = 0

y\quad =\quad 3x-1\\ \\ y-3x\quad =\quad -1\\ \\ y-3x+1\quad =\quad 0\\ \\ -3x+y+1\quad =\quad 0

Final answers,

\boxed { y\quad =\quad 3x-1 }

and 

\boxed { -3x+y+1\quad =\quad 0 }

I hope this was clear enough :)




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