Question

How do you solve `4x^2+17x+15=0`?

Answer 1

`x = -3` or `x = -5/4`

Explanation:

`4x^2 + 17x+15=0`

Do

`4x^2+12x+5x+15=0`

Then

`4x(x+3)+5(x+3)=0`

`(4x+5)(x+3)=0`

So

`x =-3" "` or `" "x=-5/4`

Answer 2

`-3, and -5/4`

Explanation:

Method 1. Use the improved quadratic formula
`y = 4x^2 + 17x + 15 = 0`.
`D = d^2 = b^2 - 4ac = 289 - 240 = 49` --> `d = +- 7`
There are 2 real roots:
`x = -b/(2a) +- d/(2a) = - 17/8 +- 7/8`
`x1 = -24/8 = - 3`
`x2 = -10/8 = -5/4`
Method 2. Use the new Transforming Method (Socratic, Google Search)
Transformed equation: `y' = x^2 + 17x + 60`.
Proceeding. Find the 2 real roots of y', then, divide them by a = 4.
Find 2 real roots, bot negative (Rule of signs), knowing the sum (-b = -17), and the product (ac = 60). They are - 5 and - 12.
The 2 real roots of y are:
`x1 = -5/a = -5/4`, and `x2 = -12/a = -12/4 = -3`

Answer 3

`x-3 or x = -5/4`

Explanation:

Factorise the quadratic trinomial.

Find factors of `4 and 15` whose products add to `17`

`" "4 and 15`
`" "darr" "darr`
`" "4" "5" "rarr 1 xx 5 =5`
`" "1" "3" "rarr 4xx3 = ul12`
`color(white)(xxxxxxxxxxxxxxxxx)17`

The signs are all positive.

`(4x+5)(x+3)=0`

Either factor can be equal to `0`

If `4x+5=0" "rarr x =-5/4`

If `x+3=0" "rarr x =-3`