Question

How do you solve `5x ^ { 2} - 17x + 6= 0`?

Answer 1

Break the trinomial down into two binomials and solve for each binomial

Explanation:

`Ax^2 + Bx + C` is the standard form of the trinomial.

The B in this equation is negative.
The C in this equation is positive.

This means that for B to be negative both factors of C must be negative.

The sum of the products of the A factors times the C factors, must equal -17.

`(5 xx 3 = 15 ) + ( 1 xx 2 = 2)`

`-15 + -2 = -17` so

`( 5x - 2) xx ( 1x -3) = 5x^2 -17x + 6` so

`5x -2 = 0` add 2 to both sides

`5x -2 +2 = 0 + 2` this gives

`5x = 2` divide both sides by 5

`5x/5 = 2/5` this gives.

`x = 2/5`

`1x - 3 = 0` add three to both sides.

`1x -3 +3 = 0 + 3` this gives.

`1x = 3`

So x equals both 2/5 and 3

Answer 2

2/5 and 5

Explanation:

`y = 5x^2 - 17x + 6 = 0`
Use the new Transforming Method (Socratic, Google Search)
Transformed equation:
`y' = x^2 - 17x + 30 = 0`
Method: find 2 real roots of y', then, divide them by a = 5.
Find 2 real roots of y', knowing sum (-b = 17) and product (ac = 30).
They are : 2 and 15.
Back to original y, the 2 real roots are:
`x1 = 2/a = 2/5`, and
`x2 = 15/a = 15/5 = 3`