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:
Step-by-step explanation:
13) x⁴-12x² +36
(a-b)² = a²-2ab+b²
a = x² ; b = 6
(x²)² - 2 * x² * 6 + 6² = (x² - 6)²
14) w⁴- 14w² - 32 = w⁴+ 2w² - 16w² - 32 = w² (w² + 2) - 16 (w²+2)
= (w² + 2) (w² -16 )
15) k³ + 7k² - 44k = k ( k² + 7k -44) = k ( k+11 ) ( k-4 )
16) 2a³ +28a²+96a =2a(a²+14a+48) = 2a(a+6)(a+8)
17) -x³ +4x² +21x = (-x) ( x² - 4x - 21) = (-x)(x-7)(x+3)
18) m⁶ - 7m⁴ -18m² = m² ( m⁴-7m²-18) = m² (m²-9)(m²+9)
= m² (m+1) (m-1)(m²+9)
19) 9y⁶ +6y⁴ + y²= y² ( 9y⁴+6y²+1) = y² (3y²+1)²
20) 8c⁴+10c² -3 = (4c +1)(2c-3)
Answer:
see below
Step-by-step explanation:
sqrt(90)
We know that sqrt(ab) = sqrt(a) sqrt(b)
sqrt(9*10)
sqrt(9) sqrt(10)
3*sqrt(10)
Your answer will be 7.50. What I did was take 1.25 and times it by 6.
1a) The function has arrows on both ends and no place in the middle where it is not defined. Its domain is ...
All Reals
1b) The function gives no output values below -3, but it gives output values of -3 and all above that. Its range is ...
y ≥ -3
1c) For values of x less than -1, the function's output is 1. This matches g(x) and s(x). At x=0, the function's output is -3, which only matches g(x). The appropriate choice is ...
g(x)
2b) The function is only defined for 0 ≤ x < 8. This is its domain.
3) A definition might be ...