Algebra PLZ HELP

Question

11. Solve by quadratic formula: (on your resource page)
a. 4x2 + 3x – 8 = 0 b. 2x2 + 7x = 6


12. Using the discriminant, how many real number solutions does this equation have?
3x2 – 2 = 5x

13. A ball is thrown into the air. The height in feet of the ball can be modeled by the equation h = -16t2 + 20t + 6 where t is the time in seconds, the ball is in the air. When will the ball hit the ground? How high will the ball go?


14. Use the Zero Product Property to solve the equation:
-3m(2m – 5) = 0
15. What are the solutions to the system? (Hint: graphing is the easiest!)
y = x2 + 2x – 45
y = -3x + 5

Answer 1

We can do this in general for arbitrary quadratic coefficients A,B,C to produce the quadratic formula:

 

Ax^2+Bx+C=0

x^2+(B/A)x = -C/A

 

We next complete the square on the LHS by adding B^2/4A^2 to both sides and recognizing that 

(x+b)^2=x^2+2bx+b^2...so with b=B/2A...we have

 

(x+B/2A)^2 = B^2/4A^2 - C/A

 

we can combine the RHS with a common denominator to get:

 

(x+B/2A)^2 = (B^2-4AC)/4A^2

 

Now apply a square root and subtract B/2A from both sides to produce the formula:

 

x = [-B +/- \sqrt(B^2-4AC)]/2A

 

In a. we can substitute the values A=4, B=3, C=-8 and find the solutions for x...in b. we can first subtract 6 from both sides and then identify the coefficients of our quadratic that is equal to 0 and apply the same quadratic formula for these new coefficients.

 

12.  The discriminant is the terms inside the \sqrt() in the Q.F.  When this is positive, we have 2 real solutions...when it is 0 we have one real solution...and when it is negative we get no real solutions...

Subtract 5x from both sides, identify the coefficients you have here in the form Ax^2+Bx+C=0, and compute the discriminant: B^2-4AC...compare this to 0 to find the number of real solutions!

 

13. When the ball hits the ground the height is 0...so solve for the time, t when h=0=-16t^2+20t+6 (hint use the Q.F.)

Doing this should produce two solutions...one with a positive value for t and one with a negative value...Since time cannot be negative (time travel backwards is theoretically impossible), the positive root is the time when the ball hits the ground.  To find the maximal height the ball reaches, we can use our knowledge of quadratics to say that the maximal height occurs at the point midway between the two roots...so adding our two solution values of t (from the QF) together and dividing by 2 will yield the time when it reaches a maximum, and plugging this avg value of the roots into the original equation for h will produce the maximal height!

 

14. Zero product property says that if we multiply two things and get 0, one or the other (or both) must be 0!

If -3m=0, what is m?

If 2m-5=0, what is m?

 

15. Solutions of a system can be found by looking at the intersection of the two graphs.

Algebraically we can solve a system by substitution or elimination.

Substitution says use one formula (say for y in this case) to rewrite the other equation:

Since y=-3x+5, we get -3x+5=x^2+2x-45 in the first...we can solve this by subtracting the LHS and using the QF.

Elimination method would be to subtract the entire second equation from the first:

 

 (y = x^2+2x–45)
-(y =0x^2-3x + 5)

0=x^2+5x-50

Now we can apply the QF and find the solutions for x...plugging these (possibly 2 of them) values back into EITHER equation, will yield the corresponding values for y...this will identify the points in the plane where the 2 graphs intersect!

 

Hope that helps you!