Question

if alpha , beta are the roots of x^2-5x-4=0, then find the quadratic equation whose roots are (alpha+2)/3 , (beta+2)/3​

Answer 1

Answer:

Step-by-step explanation:

We know in a quadratic equation of form ax² + bx + c = 0, sum of roots = -b/a and product of roots = c/a.

So in equation, x² - 5x - 4 = 0, whose roots are  α and β

a = 1 , b = -5, c = -4

=> Sum of roots = α + β = -(-5)/1 = 5.

    product of roots = αβ = c/a = -4/1 = -4.

For the new equation, roots are given as (α + 2)/3, (β + 2)/3. Also

The new quadratic equation will be of form x² + (Sum of roots)x + (product of roots) = 0.

Sum of roots = (α + 2)/3 + (β + 2)/3

                      = (α + β + 4) / 3

                      = (5 + 4)/3  (Since α + β = 5)

                      = 9/3

                      = 3.

Product of roots = (α + 2)/3  * (β + 2)/3

                           = 1/9[ (α + 2)  (β + 2)]

                           = 1/9[ αβ  + 2α + 2β + 4]

                           = 1/9[-4+ 2(5)+4] (Since αβ = -4 and α + β = 5)

                            = 10/9

Now substituting the values, we get x² + (3)x + 10/9 = 0

                  = > 9x² + 27x + 10 = 0.