anonymous
  • anonymous
Please help...Use the quadratic and a calculator to solve the following equation. Round intermediate calculations and final answers to two decimal places. 6.25y^2 + 12y + 5.76 = 0 i got y = .96, -96 but it was wrong should i round up to 1, -1
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
if you require the answer correct to dec places to do not iround intermediate results to 2 dp's
anonymous
  • anonymous
right so I dont understand the statement Round intermediate calculations and final answers to two decimal places
anonymous
  • anonymous
2 dec places i meant to say

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anonymous
  • anonymous
Is it bcuz it is saying Use the quadratic and a calculator to solve
anonymous
  • anonymous
i make the solutios -0.96
anonymous
  • anonymous
there is only one solution
anonymous
  • anonymous
so you think incluce the 0
anonymous
  • anonymous
can a solution set have only one thing in it?
anonymous
  • anonymous
yes
anonymous
  • anonymous
6.25y^2 + 12y + 5.76 = 0 Rationalize the coefficients. \[\frac{25}{4}y^2+12y+\frac{144}{25}=0\]Multiply each side by 100 and simplify.\[625 y^2+1200 y+576=0 \]Use the binomial theorem or, to pull a rabbit out of the hat:\[625 y^2+1200 y+576=(25 y+24)^2 \]\[y=\frac{24}{25}\text{ }\text{or}\text{ }0.96 \]
anonymous
  • anonymous
if you graph this function it will just touch the x axis at -0.96
anonymous
  • anonymous
I stand corrected. Yes, the sign is negative. Thank you.
anonymous
  • anonymous
ok
anonymous
  • anonymous
so you are saying that the only solution in the solution set is -0.96??
anonymous
  • anonymous
There are two solutions, each with identical values.
anonymous
  • anonymous
so both would be -0.96
anonymous
  • anonymous
yes
anonymous
  • anonymous
but I probably only need to enter it once?
anonymous
  • anonymous
the books always say two identical solutions but to be honest i dont understand why
anonymous
  • anonymous
Woohoo it was correct...thank you both sooooo much!!!
anonymous
  • anonymous
I use Mathematica to do the calculations.\[\text{Solve}\left[(25 y+24)^2\text{==}0,y\right]=\left\{\left\{y\to -\frac{24}{25}\right\},\left\{y\to -\frac{24}{25}\right\}\right\} \]Notice that it returned two values for y. I would put down as an answer, x = -0.96 twice. Does this answer you questions?
anonymous
  • anonymous
no probs popcorn
anonymous
  • anonymous
Your welcome.

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