How many real number solutions exist for 2x^2 + 8x + 8 = 0
The first part is 2x to the power of 2.
2x^2 + 8x + 8 = 0
x^2 + 4x + 4 = 0
discriminant = b^2 - 4ac
= 16 - 4(1)(4)= 0
so there is 1 real answer
check:
x^2 + 4x + 4 = 0
(x+2)^2 = 0
x+2=0
x = -2
To find the number of real number solutions for the equation 2x^2 + 8x + 8 = 0, we can use the quadratic formula.
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by x = (-b ± √(b^2 - 4ac))/(2a).
Comparing this with the given equation, we have a = 2, b = 8, and c = 8.
Substituting these values into the quadratic formula, we have:
x = (-8 ± √(8^2 - 4*2*8))/(2*2)
x = (-8 ± √(64 - 64))/(4)
x = (-8 ± √(0))/(4)
Since the discriminant (b^2 - 4ac) is zero, we have √(0) which simplifies to 0. Therefore, the solutions for x become:
x = (-8 ± 0)/(4)
x = -8/4
x = -2
So the equation 2x^2 + 8x + 8 = 0 has only one real number solution, which is x = -2.
To find the number of real number solutions for the equation 2x^2 + 8x + 8 = 0, we can use the discriminant of the quadratic equation. The discriminant is calculated using the formula D = b^2 - 4ac, where a, b, and c are the coefficients of the equation in standard form ax^2 + bx + c = 0.
For our equation 2x^2 + 8x + 8 = 0, the coefficients are a = 2, b = 8, and c = 8. Substituting these values into the discriminant formula, we get D = (8)^2 - 4(2)(8) = 64 - 64 = 0.
The value of the discriminant D = 0 indicates that the equation has exactly one real number solution. Furthermore, if the discriminant is positive (D > 0), the equation would have two distinct real number solutions. If the discriminant is negative (D < 0), the equation would have no real number solutions but two complex solutions.
Therefore, the equation 2x^2 + 8x + 8 = 0 has exactly one real number solution.