I'm pretty sure it has to have 1 over any number
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](https://tex.z-dn.net/?f=%20x%5E2%20%2B%2011x%20-%204%20%3D%200%20)
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}](https://tex.z-dn.net/?f=%20x_%7B1%2C2%7D%20%3D%20%5Cfrac%7B-b%5Cpm%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2%7D%20)
where
is a compact way to indicate both solutions
and
, while
are the coefficients of the quadratic equation, i.e. we consider the polynomial
.
So, in your case, we have ![a=1,\ \ b=11,\ \ c=-4](https://tex.z-dn.net/?f=%20a%3D1%2C%5C%20%5C%20b%3D11%2C%5C%20%5C%20c%3D-4%20)
Plug those values into the formula to get
![x_{1,2} = \frac{-11\pm\sqrt{121+16}}{2} = \frac{-11\pm\sqrt{137}}{2}](https://tex.z-dn.net/?f=%20x_%7B1%2C2%7D%20%3D%20%5Cfrac%7B-11%5Cpm%5Csqrt%7B121%2B16%7D%7D%7B2%7D%20%3D%20%5Cfrac%7B-11%5Cpm%5Csqrt%7B137%7D%7D%7B2%7D%20)
So, the two solutions are
![x_1 = \frac{-11+\sqrt{137}}{2}](https://tex.z-dn.net/?f=%20x_1%20%3D%20%5Cfrac%7B-11%2B%5Csqrt%7B137%7D%7D%7B2%7D%20)
![x_2 = \frac{-11-\sqrt{137}}{2}](https://tex.z-dn.net/?f=%20x_2%20%3D%20%5Cfrac%7B-11-%5Csqrt%7B137%7D%7D%7B2%7D%20)
Times the bottom equation by -4 to cancel out the x’s.
Add the equations.
Solve for y.
Plug in y to find x.
Answer:
17 is the answer.
Step-by-step explanation:
Question is not complete, so i have attached it.
Answer:
B: -2x² - 6x² - 6x + 3x + 3 + 2
Step-by-step explanation:
From the image attached, the given polynomial is;
3 - 6x - 2x² + 3x - 6x² + 2
For like terms to be next to each other, we have to rearrange it as follows;
-2x² - 6x² - 6x + 3x + 3 + 2
Like terms are x² and x, thus they are now arranged next to each other.
The only option that corresponds to our answer is option B