X² + 1 = 0
=> (x+1)² - 2x = 0
=> x+1 = √(2x)
or x - √(2x) + 1 = 0
Now take y=√x
So, the equation changes to
y² - y√2 + 1 = 0
By quadratic formula, we get:-
y = [√2 ± √(2–4)]/2
or √x = (√2 ± i√2)/2 or (1 ± i)/√2 [by cancelling the √2 in numerator and denominator and ‘i' is a imaginary number with value √(-1)]
or x = [(1 ± i)²]/2
So roots are [(1+i)²]/2 and [(1 - i)²]/2
Thus we got two roots but in complex plane. If you put this values in the formula for formation of quadratic equation, that is x²+(a+b)x - ab where a and b are roots of the equation, you will get the equation
x² + 1 = 0 back again
So it’s x=1 or x=-1
Answer= 120
Explanation= plugged it into a calculator
Given:
The tens digit of a two digit number is 5 greater the units digit.
If you subtract double the reversed number from it, the result is a fourth of the original number.
To find:
The original number.
Solution:
Let n be the two digit number and x be the unit digit. Then tens digit is (x+5) and the original number is:



Reversed number is:


If you subtract double the reversed number from it, the result is a fourth of the original number.



Multiply both sides by 4.



Divide both sides by 55.


The unit digit is 2. So, the tens digit is
.
Therefore, the original number is 72.
Step-by-step explanation:Step 1: Simplify both sides of the equation.
4−(2y−1)=2(5y+9)+y
4+−1(2y−1)=2(5y+9)+y(Distribute the Negative Sign)
4+−1(2y)+(−1)(−1)=2(5y+9)+y
4+−2y+1=2(5y+9)+y
4+−2y+1=(2)(5y)+(2)(9)+y(Distribute)
4+−2y+1=10y+18+y
(−2y)+(4+1)=(10y+y)+(18)(Combine Like Terms)
−2y+5=11y+18
−2y+5=11y+18
Step 2: Subtract 11y from both sides.
−2y+5−11y=11y+18−11y
−13y+5=18
Step 3: Subtract 5 from both sides.
−13y+5−5=18−5
−13y=13
Step 4: Divide both sides by -13.
−13y
−13
=
13
−13
y=−1
Answer: y =-1