[9.02] Identify the solutions of 3x2 − x − 4 = 0. x = −1 over 3, x = 4 x = 1 over 3, x = −4 x = 4 over 3?

Question

[9.02] Identify the solutions of 3x2 − x − 4 = 0.

x = −1 over 3, x = 4
x = 1 over 3, x = −4
x = 4 over 3, x = −1
x = −4 over 3, x = 1
16. [9.02] Identify the solutions of 3x2 − 7x + 4 = 0.

x = −1 over 3, x = −4
x = 1 over 3, x = 4
x = −4 over 3, x = −1
x = 4 over 3, x = 1
17. [9.05] How many and of what type are the solutions of a quadratic equation when the value of the radicand is 25?

No real solutions
Two identical rational solutions
Two different rational solutions
Two irrational solutions
18. [9.05] Identify the graph of a quadratic equation with two irrational solutions.

u-shaped graph opening up and crossing the x-axis at approximately (−2.41, 0) and (0.41, 0)
u-shaped graph opening up and crossing the axis at (−3, 0) and (1, 0)
u-shaped graph opening down and touching the axis at (−2, 0)
u-shaped graph opening down and not crossing the axis.
19. [9.05] How many and of what type are the solutions to x2 + 2x + 4 = 0?

No real solutions
Two identical rational solutions
Two different rational solutions
Two irrational solutions
20. [9.05] What are the exact solutions of x2 − 3x − 1 = 0?

x = the quantity of negative 3 plus or minus the square root of 13 all over 2
x = the quantity of 3 plus or minus the square root of 13 all over 2
x = the quantity of negative 3 plus or minus the square root of 5 all over 2
x = the quantity of 3 plus or minus the square root of 5 all over 2
21. [9.05] What are the exact solutions of x2 + 5x + 2

Answer 1

x=((1+/-√1+48))6=(1+/-7)/6= 4/3,-1

#16 is all you get from me, cause I'm not taking the course, U R. Do the rest yourself !

Answer 2

3x^2 - x - 4 = (3x -4)(x+1) = 0

==> x = 4/3 or x = -1 (the third choice)

3x^2 -7x + 4 = (3x-4)(x-1)= 0

==> x = 4/3 or x = 1 (the last choice)

....

Now, "radicand"... I guess you're talking about the general solution to a quadratic:

so, if ax^2 + bx + c = 0 then x = (-b +/- √(b^2 - 4ac))/2a

The radicand would then be (b^2 - 4ac) then use the discriminant rule:

if < 0 then two complex solutions

if = 0 then one "double root"

if > 0 then two real roots.

Since √25 = 5, the roots would be real and rational.

Answer 3

16. [9.02] None of the proposed answers are correct. It should be:

x = 4 / 3, x = −1



17. [9.05] Two different rational solutions



18. [9.05] u-shaped graph opening up and crossing the x-axis at approximately (−2.41, 0) and (0.41, 0)



19. [9.05] No real solutions



20. [9.05] x = the quantity of 3 plus or minus the square root of 13 all over 2



21. [9.05] x = (-5 + √17) / 2 , x = (-5 - √17) / 2