Question

Solve 3x^2 + 17x - 6 = 0.

3x2 + 18x - x- 6 =0

3x(x+6) -1(x+6) =0

(3x-1)(x+6) =0

3x-1 or x+6 =0


Based on the work shown above, which of these values are possible solutions of the equation? Check all of the boxes that apply.


x=-6

x=6

x=-⅓

x=⅓

x=0

Answer 1

3x^2+17x-6=0.

a = 3; b = 17; c = -6;

Δ = b2-4ac

Δ = 172-4·3·(-6)

Δ = 361

The delta value is higher than zero, so the equation has two solutions

We use following formulas to calculate our solutions:

x1=−b−Δ√2ax2=−b+Δ√2a

Δ−−√=361−−−√=19

x1=−b−Δ√2a=−(17)−192∗3=−366=−6

x2=−b+Δ√2a=−(17)+192∗3=26=1/3

Explanation:

Answer 2

-6 and 1/3

Explanation: