Answer:
The answer is 22458786952
The solution of
are 1 + 2i and 1 – 2i
<u>Solution:</u>
Given, equation is ![x^{2}-2 x+5=0](https://tex.z-dn.net/?f=x%5E%7B2%7D-2%20x%2B5%3D0)
We have to find the roots of the given quadratic equation
Now, let us use the quadratic formula
--- (1)
<em><u>Let us determine the nature of roots:</u></em>
Here in
a = 1 ; b = -2 ; c = 5
![b^2 - 4ac = 2^2 - 4(1)(5) = 4 - 20 = -16](https://tex.z-dn.net/?f=b%5E2%20-%204ac%20%3D%202%5E2%20-%204%281%29%285%29%20%3D%204%20-%2020%20%3D%20-16)
Since
, the roots obtained will be complex conjugates.
Now plug in values in eqn 1, we get,
![x=\frac{-(-2) \pm \sqrt{(-2)^{2}-4 \times 1 \times 5}}{2 \times 1}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-%28-2%29%20%5Cpm%20%5Csqrt%7B%28-2%29%5E%7B2%7D-4%20%5Ctimes%201%20%5Ctimes%205%7D%7D%7B2%20%5Ctimes%201%7D)
On solving we get,
![x=\frac{2 \pm \sqrt{4-20}}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B2%20%5Cpm%20%5Csqrt%7B4-20%7D%7D%7B2%7D)
![x=\frac{2 \pm \sqrt{-16}}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B2%20%5Cpm%20%5Csqrt%7B-16%7D%7D%7B2%7D)
![x=\frac{2 \pm \sqrt{16} \times \sqrt{-1}}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B2%20%5Cpm%20%5Csqrt%7B16%7D%20%5Ctimes%20%5Csqrt%7B-1%7D%7D%7B2%7D)
we know that square root of -1 is "i" which is a complex number
![\begin{array}{l}{\mathrm{x}=\frac{2 \pm 4 i}{2}} \\\\ {\mathrm{x}=1 \pm 2 i}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B%5Cmathrm%7Bx%7D%3D%5Cfrac%7B2%20%5Cpm%204%20i%7D%7B2%7D%7D%20%5C%5C%5C%5C%20%7B%5Cmathrm%7Bx%7D%3D1%20%5Cpm%202%20i%7D%5Cend%7Barray%7D)
Hence, the roots of the given quadratic equation are 1 + 2i and 1 – 2i
The correct result would be y = -6/5 - (2 * sqrt(6))/5 and<span> y = (2 * sqrt(6))/5 - 6/5.
</span>
Correct question:
a sequence U1, U2,U3...... is defined by the relation
. if U1 =2, find U2 and U3
Answer:
U2 = 9
U3 = 23
Step-by-step explanation:
Given sequence;
![U_{n +1} = 2U_n +5](https://tex.z-dn.net/?f=U_%7Bn%20%2B1%7D%20%3D%202U_n%20%2B5)
U1 = 2
U2 = ?
![U_{n +1} = 2U_n +5\\\\U_2 = 2(U_1) + 5\\\\U_2 = 2(2) + 5\\\\U_2 = 4 + 5\\\\U_2 = 9](https://tex.z-dn.net/?f=U_%7Bn%20%2B1%7D%20%3D%202U_n%20%2B5%5C%5C%5C%5CU_2%20%3D%202%28U_1%29%20%2B%205%5C%5C%5C%5CU_2%20%3D%202%282%29%20%2B%205%5C%5C%5C%5CU_2%20%3D%204%20%2B%205%5C%5C%5C%5CU_2%20%3D%209)
Solving for U3;
![U_3 = 2(U_2) + 5\\\\U_3 = 2(9) + 5\\\\U_3 = 18 + 5\\\\U_3 = 23](https://tex.z-dn.net/?f=U_3%20%3D%202%28U_2%29%20%2B%205%5C%5C%5C%5CU_3%20%3D%202%289%29%20%2B%205%5C%5C%5C%5CU_3%20%3D%2018%20%2B%205%5C%5C%5C%5CU_3%20%3D%2023)
Answer:
Option 2.
Step-by-step explanation:
X is a number greater than the equality. Therefore the arrow will go to the bigger side. It also is not equal to the equality hence the cirle is open.