Answer:
This quadratic equation has 2 solutions.
Step-by-step explanation:
I assume the '?' in your question is meant to be power 2 (²), or else it would not be a quadratic equation. You could write it using the superscript version of 2.
We can solve this equation by expressing it in the form: ax² + bx + c
x² + 9x= -8
x² + 9x + 8 = 0
Now if you know the discriminant, you can simply plug in your values of a, b, and c to see how many solutions there are.
In this case, you would not need the discriminant as there are whole-number factors and hence this can simply be factorised.
x² + 9x + 8 = 0
(x + 8)(x + 1) = 0
For this equation to be true (= 0), x can equal -8 OR -1.
Hence, this quadratic equation has 2 solutions.
![\bf [sin(x)+cos(x)]^2+[sin(x)-cos(x)]^2=2 \\\\\\\ [sin^2(x)+2sin(x)cos(x)+cos^2(x)]\\\\+~ [sin^2(x)-2sin(x)cos(x)+cos^2(x)] \\\\\\ 2sin^2(x)+2cos^2(x)\implies 2[sin^2(x)+cos^2(x)]\implies 2[1]\implies 2](https://tex.z-dn.net/?f=%5Cbf%20%5Bsin%28x%29%2Bcos%28x%29%5D%5E2%2B%5Bsin%28x%29-cos%28x%29%5D%5E2%3D2%0A%5C%5C%5C%5C%5C%5C%5C%0A%5Bsin%5E2%28x%29%2B2sin%28x%29cos%28x%29%2Bcos%5E2%28x%29%5D%5C%5C%5C%5C%2B~%20%5Bsin%5E2%28x%29-2sin%28x%29cos%28x%29%2Bcos%5E2%28x%29%5D%0A%5C%5C%5C%5C%5C%5C%0A2sin%5E2%28x%29%2B2cos%5E2%28x%29%5Cimplies%202%5Bsin%5E2%28x%29%2Bcos%5E2%28x%29%5D%5Cimplies%202%5B1%5D%5Cimplies%202)
![\bf \cfrac{2-cos^2(x)}{sin(x)}=csc(x)+sin(x) \\\\\\ \cfrac{2-[1-sin^2(x)]}{sin(x)}\implies \cfrac{2-1+sin^2(x)}{sin(x)}\implies \cfrac{1+sin^2(x)}{sin(x)} \\\\\\ \cfrac{1}{sin(x)}+\cfrac{sin^2(x)}{sin(x)}\implies csc(x)+sin(x)](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B2-cos%5E2%28x%29%7D%7Bsin%28x%29%7D%3Dcsc%28x%29%2Bsin%28x%29%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B2-%5B1-sin%5E2%28x%29%5D%7D%7Bsin%28x%29%7D%5Cimplies%20%5Ccfrac%7B2-1%2Bsin%5E2%28x%29%7D%7Bsin%28x%29%7D%5Cimplies%20%5Ccfrac%7B1%2Bsin%5E2%28x%29%7D%7Bsin%28x%29%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B1%7D%7Bsin%28x%29%7D%2B%5Ccfrac%7Bsin%5E2%28x%29%7D%7Bsin%28x%29%7D%5Cimplies%20csc%28x%29%2Bsin%28x%29)
