Find all real roots of the following equation.
x^2 - (5/3)x = -2/3
Thx
first get rid of nasty fractions by multiplying each term by 3 , LCD
3x^2 - 5x + 2 = 0
which factors quite nicely.
Let me know what you got
Thx. After getting rid of the fractions, I could easily use the quadratic formula to find the roots, (1, 2/3)! :D
good job
To find the real roots of the equation x^2 - (5/3)x = -2/3, we need to solve for the values of x that satisfy the equation.
Step 1: Move all terms to one side of the equation to get a quadratic equation in standard form:
x^2 - (5/3)x + 2/3 = 0
Step 2: We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Since the equation can easily be factored, let's proceed with factoring:
(x - a)(x - b) = 0
To find the values of a and b, we need to find two numbers whose product is 2/3 and whose sum is -5/3. After trying different combinations, we can see that -2/3 and -1/3 satisfy these conditions, since (-2/3) * (-1/3) = 2/3 and (-2/3) + (-1/3) = -5/3.
Therefore, we can factor the quadratic equation as follows:
(x - 2/3)(x - 1/3) = 0
Step 3: Now, set each factor equal to zero and solve for x:
x - 2/3 = 0 or x - 1/3 = 0
Solving for x in each equation, we get:
x = 2/3 or x = 1/3
So, the real roots of the equation x^2 - (5/3)x = -2/3 are x = 2/3 and x = 1/3.