anonymous
  • anonymous
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)
Mathematics
katieb
  • katieb
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freckles
  • freckles
So part A sounds like it i asking you to find the discriminant
freckles
  • freckles
\[ax^2+bx+c=0 \\ \text{ then the discriminant is } b^2-4ac\]
anonymous
  • anonymous
I already have part A I just need Part B.

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freckles
  • freckles
\[\text{ if } b^2-4ac>0 \text{ then you have two real solutions } \\ \text{ if } b^2-4ac=0 \text{ then you have one real solution (multiplicity 2)} \\ \text{ if } b^2-4ac<0 \text{ then you have two complex solutions }\] (assuming of course that a,b,c are real coefficients)
welshfella
  • welshfella
for part B you can use factorisation
freckles
  • freckles
Well you can use quadratic formula. Or completing the square.
freckles
  • freckles
Or factoring.
anonymous
  • anonymous
so it would be 1
anonymous
  • anonymous
freckles
  • freckles
that would be one solution
freckles
  • freckles
there is another solution
freckles
  • freckles
And we are talking about 2x^2-9x+7=0 right?
anonymous
  • anonymous
yes
freckles
  • freckles
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 ( )
freckles
  • freckles
you know that x times what is equal to 2x^2? and that -1 times what is equal to 7?
anonymous
  • anonymous
-7
freckles
  • freckles
yes but what about the other question x times what gives you 2x^2
anonymous
  • anonymous
1 Attachment
freckles
  • freckles
\[x \cdot 2x=2x^2 \\ -1 \cdot (-7)=7 \\ \text{ and the middle term will be } -2x-7x \text{ which is indeed } -9x\] \[(x-1)(2x-7)=2x^2-9x+7 \\ \\ \text{ so you have that the equation } 2x^2-9x+7=0 \\ \text{ is equivalent to the equation } (x-1)(2x-7)=0 \\ \text{ which is equivalent to solving the conjunction } x-1=0 \text{ or } 2x-7=0\]

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