Question

A quadratic equation is shown below:

9x2 − 36x + 36 = 0

Part A: Describe the solution(s) to the equation by just determining the radicand. Show your work. (5 points)

Part B: Solve 2x2 − 9x + 7 = 0 using an appropriate method. Show the steps of your work and explain why you chose the method used. (5 points)

Answer 1

So part A sounds like it i asking you to find the discriminant

Answer 2

\[ax^2+bx+c=0 \\ \text{ then the discriminant is } b^2-4ac\]

Answer 3

I already have part A I just need Part B.

Answer 4

\[\text{ if } b^2-4ac>0 \text{ then you have two real solutions } \\ \text{ if } b^2-4ac=0 \text{ then you have one real solution (multiplicity 2)} \\ \text{ if } b^2-4ac<0 \text{ then you have two complex solutions }\]
(assuming of course that a,b,c are real coefficients)

Answer 5

for part B you can use factorisation

Answer 6

Well you can use quadratic formula.
Or completing the square.

Answer 7

Or factoring.

Answer 8

so it would be 1

Answer 9

@freckles

Answer 10

that would be one solution

Answer 11

there is another solution

Answer 12

And we are talking about 2x^2-9x+7=0 right?

Answer 13

yes

Answer 14

so if you figured out x=1 is solution then you know that x-1 is a factor

(x-1)( )=2x^2-9x+7

what goes in then second pair of ( )

Answer 15

you know that x times what is equal to 2x^2?
and that -1 times what is equal to 7?

Answer 16

-7

Answer 17

yes but what about the other question x times what gives you 2x^2

Answer 18

\[x \cdot 2x=2x^2 \\ -1 \cdot (-7)=7 \\ \text{ and the middle term will be } -2x-7x \text{ which is indeed } -9x\]

\[(x-1)(2x-7)=2x^2-9x+7 \\ \\ \text{ so you have that the equation } 2x^2-9x+7=0 \\ \text{ is equivalent to the equation } (x-1)(2x-7)=0 \\ \text{ which is equivalent to solving the conjunction } x-1=0 \text{ or } 2x-7=0\]