The complex solution of a quadratic equation are (- 2 + i ) and
(- 2 - i ).
What is Quadratic equation?
An algebraic equation of the second degree is called a quadratic equation.
Given that;
A quadratic equation is;
3x² = -12x - 15
Now, The equation is written as;
3x² + 12x + 15 = 0
Take 3 common, we get;
3 (x² + 4x + 5) = 0
x² + 4x + 5 = 0
Factorize the equation by using Sridharacharya Formula;
x = - 4 ± √4² - 4*1*5 / 2*1
x = -4 ± √16 - 20 / 2
x = - 4 ± √-4 / 2
Since, √-1 = i
x = -4 ± 2i / 2
x = - 2 ± i
It gives two values of x as;
x = - 2 + i
And, x = - 2 - i
Hence, The complex solution of a quadratic equation are (- 2 + i ) and
(- 2 - i ).
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Answer:
-1/5n + 7 = 2
Start by subtracting 7 from both sides:
-1/5n + 7 - (7) = 2 - (7)
-1/5n = -5
Multiply both sides by the reciprocal of -1/5, or -5.
(-5) · (-1/5n) = (-5) · (-5)
n = 25
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Answer:
32
Step-by-step explanation:
n(A∪B)=n(A)+n(B)−n(A∩B)−−−−−−−(1)
Given n(A)= ? we represent with x
n(B)= 16
n(A∪B) = 32
Substituting in equation 1 to get n(A)
32 = n(A) + 9 − 9
⇒n(A) = 32 − 0
n(A) = 32
to confirm this we put the values into the formula below
n(A∪B)=n(A)+n(B)−n(A∩B)−−−−−−−(1)
32 = 32 + 9 - 9
Answer:
y=-2x+2
Step-by-step explanation:
If you bring the 4x to the other side by subtracting it, you will have 2y=-4x+8
then divide by the 2 to get y=-2x=4
then you substitute the 5 in for y and the -1 in for x. multiply and then subtract
