How would you solve 3x^2+8x-91=0 and 9r^2+4r+2=0 using the quadratic formula?
the same way you solve any quadratic. The first is:
x = [-8±√(8^2-4(3)(-91))]/(2*3)
= (-8±√1156)/6
= (-8±34)/6
= -7,13/3
solve the other in like wise.
To solve the equations using the quadratic formula, we first need to understand the quadratic formula itself. The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a, b, and c are the coefficients of the equation in the form ax^2 + bx + c = 0. Using this formula, we can solve both equations.
For the equation 3x^2 + 8x - 91 = 0:
a = 3, b = 8, c = -91
Substituting these values into the quadratic formula, we have:
x = (-8 ± √(8^2 - 4 * 3 * -91)) / (2 * 3)
Simplifying further:
x = (-8 ± √(64 + 1092)) / 6
x = (-8 ± √1156) / 6
Now, we calculate the square root of 1156:
√1156 = 34
Thus, the two solutions for this equation are:
x1 = (-8 + 34) / 6 = 26/6 = 13/3
x2 = (-8 - 34) / 6 = -42/6 = -7
For the equation 9r^2 + 4r + 2 = 0:
a = 9, b = 4, c = 2
Substituting these values into the quadratic formula, we have:
r = (-4 ± √(4^2 - 4 * 9 * 2)) / (2 * 9)
r = (-4 ± √(16 - 72)) / 18
r = (-4 ± √(-56)) / 18
As the square root of a negative number is not real, this equation does not have any real solutions. It has complex solutions which involve imaginary numbers.