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:
hi
Step-by-step explanation:
the area of the front face by the width. Vc= 20 cm2 x 10 cm. = 200 cm. Area = 20 cm? 2 ft. The volume of soil in the planter is 36 cubic feet. 6 ft. I draw a rectangular prism and I have a prism with the dimensions of 8 in by 12 in by 20 in.
hope this helps
Answer:The equation x² + 7 = 0 has no solution
Explanation:1- using graph:To solve the equation means graphically means to find the x-intercepts.
The attached image shows the graph of the given function.
We can note that there are no x-intercepts. This means that the given function has no real solutions
2- using algebra:To solve the equation algebraically means to find the values of x that would make the equation equal to zero.
Solving the given equation, we would find that:
x² + 7 = 0
x² = -7
x = <span>± </span>√-7
The square root of a negative number will always give imaginary values. This means that the equation has no real solutions
Hope this helps :)
Answer:
a. Tax liability using Tax Tables $8,579
b. Tax liability using Tax Rate Schedule $8,576.80
Step-by-step explanation:
Answer:
A relation is a subset of cartesian product of two non empty sets whereas A function is a type of relation in which every element of first set has one and only image in the second set.
In a relation an element of the first set can have many images in the second set whereas in a function the first element can have only one image in the second set.
The given relation is not a function as the element 1 is related to 3 different elements in the second set.
Domain={1}
Range={7,14,21}