use the quotient rule then after using it the first time use it again on your resulting equation, since you are looking for the second derivative
A. 2 (2)(3)(3)(a)(a)(a)(a)(a)(a)(a)(a)3(3)(5)(5)(a)(a)
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The equation that has the solution is 3x^2 - 10x + 6 = 0
The solution is given as:
The solution to a quadratic equation is
By comparing both equations, we have:
-b = 5
b^2 - 4ac = 7
2a = 3
Solve for b in -b = 5
b = -5
Solve for a in 2a = 3
a = 1.5
Substitute values for a and b in b^2 - 4ac = 7
(-5)^2 - 4 * 1.5c = 7
Evaluate
25 - 6c = 7
Subtract 25 from both sides
-6c = -18
Divide by - 6
c = 3
So, we have:
a = 1.5
b = -5
c = 3
A quadratic equation is represented as:
ax^2 + bx + c = 0
So, we have:
1.5x^2 - 5x +3 = 0
Multiply through by 2
3x^2 - 10x + 6 = 0
Hence, the equation that has the solution is 3x^2 - 10x + 6 = 0
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