Question

-3x^2+6x+1=4 find the discriminant of each quadratic equation then state the number and type solutions

Answer 1

Answer:

Discriminant = 0;  1 real root

Step-by-step explanation:

-3x²+6x+1=4  (subtract 4 from each side)

-3x²+6x+1 - 4 = 0

-3x²+6x-3 = 0  (factorize 3 out of the left side)

3(-x²+2x-1) = 0

-x²+2x-1 = 0

Recall that for a quadratic equation in the form:

Ax² + Bx + C = 0, the discriminant = B²-4AC

in our case, A = -1, B = 2, = -1

discriminant = B²-4AC

= 2² - 4 (-1)(-1)

= 4 - 4(1)

= 4 - 4

= 0

Recall that when discriminant = 0, the quadratic equation has 1 real root. (answer)

Answer 2

Answer:

S={( 1 )}

Step-by-step explanation:

-3x²+6x+1=4

-3x²+6x+1-4=0

-3x²+6x-3=0

∆=b²-4.a.c

∆=(6)²-4.(-3).(-3)

∆=36-36

∆=0

x'=x"=-b/2a=-(+6)/2.(-3)=-6/-6=1