A. the bottom has to be a four because 2 = a so 2a = 4. The three has to be negative because b = 3 and so -b = -3.
Answer:
a) The quantity negative three plus or minus i times the square root of thirty one over four.
Step-by-step explanation:
The given equation is [tex]2x^2 + 3x = -5[/tex]
This can be written as [tex]2x^2 + 3x + 5 = 0[/tex]
Let's factorize and find the solution.
This is a quadratic equation of the form [tex]ax^2 + bx + c =0 ; a\neq =[/tex]
Let's compare the given equation with the general form and identify the value of a, b and c.
a = 2, b = 3 and c = 5
Here we use quadratic formula to find the solution of this equation.
The quadratic formula x = [tex]\frac{-b +/- \sqrt{b^2 -4ac} }{2a}[/tex]
Now plug in the value a =2, b = 3 and c = 5 in the above quadratic formula, we get
x = [tex]\frac{-3 +/- \sqrt{3^2 - 4.2.5} }{2.2}[/tex]
x = [tex]\frac{-3 +/- \sqrt{-31} }{4}[/tex]
We know that √-1 = i
So, we get
x = [tex]\frac{-3 +/- i\sqrt{31} }{4}[/tex]
So the answer is a) The quantity negative three plus or minus i times the square root of thirty one over four.
