Question

What are the solutions to `5x^2+27x+10=0`?

Answer 1

`x=-5" or " x=-2/5`

Explanation:

`"factorise by 'splitting' the term in x"`

`rArr5x^2+25x+2x+10=0larr 25x+2x=27x`

`rArrcolor(red)(5x)(x+5)+color(red)(2)(x+5)=0`

`rArr(x+5)(color(red)(5x+2))=0`

`"equating each factor to zero"`

`rArrx+5=0rArrx=-5`

`5x+2=0rArrx=-2/5`

Answer 2

`- 5` and `- 2/5`

Explanation:

Solving by the new Transforming Method (Google, Socratic Search).
`y = 5x^2 + 27x + 10 = 0`
Transformed equation:
`y' = x^2 + 27x + 50 = 0`--> (ac = 50)
Proceeding: Find 2 real roots of y', then, divide them by a = 5.
Find 2 real roots knowing sum (-b = -27) and product (c = 50). They are -2 and -25.
Back to y, the 2 real roots are:
`- 2/a = - 2/5`, and `- 25/a = - 25/5 = - 5`
NOTE:
There are no factoring by grouping and no solving the 2 binomials.