how many real number solutions are there to the equation y=5x^2+2x-12
how many real number solutions are there to the equation y=3x^2+18x+27
Can somebody tell me how to get the solutions?
b^2-4ac = 4 + 20*12
positive so two real roots
b^2-4ac = 324 -4(3)(27) = 324-324 = 0
so two identical real number roots (the graph just grazes the x axis)
ok I get that one
is this the answers for both of them?
I decline to answer on the grounds that you can figure that out.
To find the number of real number solutions to a quadratic equation, you can use the discriminant. The discriminant is found by calculating b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation in the form ax^2 + bx + c = 0.
For the equation y = 5x^2 + 2x - 12:
a = 5, b = 2, c = -12
Discriminant = b^2 - 4ac = (2)^2 - 4(5)(-12) = 4 + 240 = 244
Since the discriminant is positive (greater than 0), this means the equation has two distinct real number solutions.
For the equation y = 3x^2 + 18x + 27:
a = 3, b = 18, c = 27
Discriminant = b^2 - 4ac = (18)^2 - 4(3)(27) = 324 - 324 = 0
Since the discriminant is zero, there is only one real number solution. This type of equation is called a perfect square trinomial.
To find the actual solutions, you can apply the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For the first equation:
x1 = (-2 + √244) / (2*5)
x2 = (-2 - √244) / (2*5)
For the second equation:
x = -b / (2a) = -18 / (2*3) = -3
Using a calculator or simplifying further, you can find the actual decimal values of the solutions.