Find all solutions of the equation x square+3x+8 =0 and express them in the form a+b
Find input the solution with b<0
The real number a is equal
To solve the equation x^2 + 3x + 8 = 0 and express the solutions in the form a + b, you can use the quadratic formula. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solution for x is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 1, b = 3, and c = 8.
Substituting these values into the quadratic formula, we get:
x = (-(3) ± √((3)^2 - 4(1)(8))) / (2(1))
Simplifying further, we have:
x = (-3 ± √(9 - 32)) / 2
x = (-3 ± √(-23)) / 2
Now, since the discriminant (b^2 - 4ac) is negative (in this case, -23), the solutions will be complex numbers. Since you are looking for a solution with b < 0 (imaginary part), we focus on the term (√(-23)).
Taking the square root of a negative number involves using the imaginary unit, "i," defined as √(-1). So, let's rewrite √(-23) as √(23)i.
The solutions can then be expressed as:
x = (-3 ± √(23)i) / 2
Now, we can split these solutions into their real and imaginary parts:
Solution 1:
x = -3/2 + (√23/2)i
This corresponds to a = -3/2 and b = √23/2
Solution 2:
x = -3/2 - (√23/2)i
This corresponds to a = -3/2 and b = -√23/2
Thus, the solution with b < 0 is x = -3/2 - (√23/2)i, which corresponds to a = -3/2.