Question

The real roots for the equation x to the power 2 by 3 plus x to the power 1 by 3 minus 2 is equals to zero​

Answer 1

Answer:

the real roots are (-3) and (2)

Step-by-step explanation:

The problem is written as follows:

(x^2)/3 + x/3 - 2 = 0

Step 1: putting all the terms with the same denominator we get:

(x^2)/3 + x/3 - 6 = 0

Step 2: which means the entire numerator is equal to 0

(x^2) + x - 6 = 0 (I call it "E")

step 3: finding the value of "b^2 - 4*a*c" in E, for it is an equation of 2nd degree, of the form

a*x^2 + b*x + c = 0

In E, we can thus find b^2 - 4*a*c,

where a=1, b=1, c=-6

so b^2 - 4*a*c = 1^2 - 4*(1)(-6) = 25

step 4: finding the real roots

since the value of b^2 - 4*a*c is positive, then we have 2 distinct roots, (which I call them R1 and R2 respectively) defined as below:

R1 = (-b-sqrt(b^2 - 4*a*c) ) / 2*a

R2 = (-b+sqrt(b^2 - 4*a*c) ) / 2*a

b^2 - 4*a*c = 25 --> sqrt(b^2 - 4*a*c) = 5

and knowing that a=1, b=1, c-6

so,

R1 = (-1-5) / 2*1 and R2 = (-1+5) / 2*1

--> R1 = -6 / 2 and R2 = 4 / 2

therefore:

R1 = -3 and R2 = 2