For the given equation;
We shall begin by expanding the parenthesis on the left side, after which we would combine all terms on and move all of them to the left side, which shall yield a quadratic equation. Then we shall solve.
Let us begin by expanding the parenthesis;
Now that we have expanded the left side of the equation, we would have;
We shall now solve the resulting quadratic equation using the quadratic formula as follows;
![\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{Where;} \\ a=9,b=-30,c=150 \\ x=\frac{-(-30)\pm\sqrt[]{(-30)^2-4(9)(150)}}{2(9)} \\ x=\frac{30\pm\sqrt[]{900-5400}}{18} \\ x=\frac{30\pm\sqrt[]{-4500}}{18} \\ x=\frac{30\pm\sqrt[]{-900\times5}}{18} \\ x=\frac{30\pm\sqrt[]{-900}\times\sqrt[]{5}}{18} \\ x=\frac{30\pm30i\sqrt[]{5}}{18} \\ \text{Therefore;} \\ x=\frac{30+30i\sqrt[]{5}}{18},x=\frac{30-30i\sqrt[]{5}}{18} \\ \text{Divide all through by 6, and we'll have;} \\ x=\frac{5+5i\sqrt[]{5}}{3},x=\frac{5-5i\sqrt[]{5}}{3} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%5C%5C%20%5Ctext%7BWhere%3B%7D%20%5C%5C%20a%3D9%2Cb%3D-30%2Cc%3D150%20%5C%5C%20x%3D%5Cfrac%7B-%28-30%29%5Cpm%5Csqrt%5B%5D%7B%28-30%29%5E2-4%289%29%28150%29%7D%7D%7B2%289%29%7D%20%5C%5C%20x%3D%5Cfrac%7B30%5Cpm%5Csqrt%5B%5D%7B900-5400%7D%7D%7B18%7D%20%5C%5C%20x%3D%5Cfrac%7B30%5Cpm%5Csqrt%5B%5D%7B-4500%7D%7D%7B18%7D%20%5C%5C%20x%3D%5Cfrac%7B30%5Cpm%5Csqrt%5B%5D%7B-900%5Ctimes5%7D%7D%7B18%7D%20%5C%5C%20x%3D%5Cfrac%7B30%5Cpm%5Csqrt%5B%5D%7B-900%7D%5Ctimes%5Csqrt%5B%5D%7B5%7D%7D%7B18%7D%20%5C%5C%20x%3D%5Cfrac%7B30%5Cpm30i%5Csqrt%5B%5D%7B5%7D%7D%7B18%7D%20%5C%5C%20%5Ctext%7BTherefore%3B%7D%20%5C%5C%20x%3D%5Cfrac%7B30%2B30i%5Csqrt%5B%5D%7B5%7D%7D%7B18%7D%2Cx%3D%5Cfrac%7B30-30i%5Csqrt%5B%5D%7B5%7D%7D%7B18%7D%20%5C%5C%20%5Ctext%7BDivide%20all%20through%20by%206%2C%20and%20we%27ll%20have%3B%7D%20%5C%5C%20x%3D%5Cfrac%7B5%2B5i%5Csqrt%5B%5D%7B5%7D%7D%7B3%7D%2Cx%3D%5Cfrac%7B5-5i%5Csqrt%5B%5D%7B5%7D%7D%7B3%7D%20%5Cend%7Bgathered%7D)
ANSWER:
![x=\frac{5+5i\sqrt[]{5}}{3},x=\frac{5-5i\sqrt[]{5}}{3}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B5%2B5i%5Csqrt%5B%5D%7B5%7D%7D%7B3%7D%2Cx%3D%5Cfrac%7B5-5i%5Csqrt%5B%5D%7B5%7D%7D%7B3%7D)
Answer:
I do not know if there is this worksheet online which comes with answers, but you can post questions here on brainly, and I can help you clarify them!!
:>>
2x + 3y/4 = 5....multiply everything by 4
8x + 3y = 20
3y = -8x + 20
y = -8/3x + 20/3 or y = (-8x + 20) / 3
Answer:
8 cm and 9 cm
Step-by-step explanation:
In order to find out, you add both 8 and 9 together. If the sum is above/higher than the number 13, they can all be made into a triangle. 8+9 is 17 and 17 is higher than 13.
hope this helps!
Answer:
There are infinite solutions because there is always another number between any two numbers.
Step-by-step explanation:
Pick any number (including decimal) add an extra decimal after the smallest one and then you have a number between them. E.g.
1.234 and 2.987
1.2341, 1.2342, 1.2343, ... are all between them
