The system 3x-y+4+4=0 and x^2-8y-1=0 will have
A. 0 real solutions
B. 1 real solutions
C. 2 real solutions
D. an infinite number of real solutions
I suspect a typo, else why the 4+4 ?
in any case, just substitute for y and then check the discriminant of the quadratic in x.
The answer is C because (25.304,79.912) and (-1.304,0.088)
tip: to check this go on Geobro or Desmos. ;)
To determine the number of real solutions for a system of equations, we need to solve the system and see if there are any solutions. Let's start by solving the system of equations:
System 1: 3x - y + 8 = 0
System 2: x^2 - 8y - 1 = 0
For System 1, we can rearrange the equation to isolate y:
y = 3x + 8
Now, we substitute the expression for y in System 2:
x^2 - 8(3x + 8) - 1 = 0
x^2 - 24x - 64 - 1 = 0
x^2 - 24x - 65 = 0
To find the solutions to this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -24, and c = -65. Plugging these values into the quadratic formula, we have:
x = (-(-24) ± √((-24)^2 - 4(1)(-65))) / (2(1))
x = (24 ± √(576 + 260)) / 2
x = (24 ± √836) / 2
x = 12 ± √209
Since the expression beneath the square root (√209) is positive, the solutions are real. Therefore, there are two real solutions for x.
Now, we can substitute these values of x back into System 1 to find the corresponding values of y:
For x = 12 + √209:
y = 3(12 + √209) + 8
y = 36 + 3√209 + 8
y = 44 + 3√209
For x = 12 - √209:
y = 3(12 - √209) + 8
y = 36 - 3√209 + 8
y = 44 - 3√209
Hence, we have two real solutions for x and y: (12 + √209, 44 + 3√209) and (12 - √209, 44 - 3√209). Therefore, the answer is C. 2 real solutions.