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how do you solve 3x^2+2x-3=0

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3 Answers

     3x2 + 2x - 3 = 0

Since this quadratic equation is set equal to 0, we know we are looking to solve for the zeros/roots of the function. Note that these solutions are the x-intercepts (i.e., they are the points when y=0 on the graph of the function).

The easiest way to solve for the zeros of the function would be to factor the quadratic equation and then use the zero product property. This equation, however, cannot be factored. With this, we have the option solving by either completing the square or using the quadratic formula. Since a≠1, the latter would be the simpler option.

Quadratic formula:     x = (-b ± √(b2 - 4ac))/2a

Given that   a = 3 ,   b = 2 , and   c = -3 , then 

          x = (-2 ± √(22 - 4(3)(-3)))/(2)(3)

             = (-2 ± √(4 + 36))/6

             = (-2 ± √(40))/6

             = (-2 ± √(4)·√(10))/6

             = (-2 ± 2√(10))/6

             = 2(-1 ± √10)/6

             = (-1 ± √10)/3

Thus, the two solutions to this quadratic function are

     x = (-1 + √10)/3       and       x = (-1 - √10)/3

This is not a factorable quadratic - you can't split (3)x(-3) = -9 (product of the coefficient of x^2 and constant term) in to two factors whose sum is +2 (coefficient of x). Use quadratic formula:

Roots of quadratic equation: a x^2 + b x + c = 0  are given by x1 = (-b +√(b^2 - 4 a c))/(2a) and x2 = (-b -√(b^2 - 4 a c))/(2a).

Here, in the given problem: a = 3, b = 2, c = -3.

√(b^2 - 4 a c) = √(2^2 - 4(3)(-3)) = √(40) = 2√(10).

x1 = (-2 + 2√(10))/(2x3) = (-1 + √(10))/3 and x2 = (-1 - √(10))/3.

You may recall that (b^2 - 4 a c) is called discriminant - if this number is a perfect square then the quadratic is factorable (roots of quadratic equation are real rational numbers).

 

Brittany,

First, divide by three to get: x2+2x/3-1=0

Next, add 1 to both sides: x2+2x/3=1

Now the tricky part, take one half of the coefficient of x, square it , and add it to both sides. In this case, the coefficient of x is 2/3, so one half of that is 1/3 and squaring that gives 1/9: x2+2x/3+1/9=10/9

Next, factor the left hand side: (x+1/3)^2=10/9

Now to get rid of the exponent on the left hand side, you must take the square root of both sides: x+1/3=√10/√9 or x+1/3=√10/3. And because we are working with squares, there is an answer on the negative side of x+1/3=-√10/3.

Finally, subtract 1/3 from each side in each equation to get x=(√10/3)-1/3 and x=(-√10/3)-1/3

I hope this helps.