anonymous
  • anonymous
(k+5)x^2 - (2k + 3)x + (k + 1) = 0 The roots are real and equal
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@freckles
campbell_st
  • campbell_st
real and equal means the discriminant is zero and the discriminant is found using \[\Delta = b^2 - 4ac\]
campbell_st
  • campbell_st
you have a = (k + 5) b = -(2k + 3) c = (k + 1) so substitute the values then you can solve for k hope it helps

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anonymous
  • anonymous
4k^2+9+12k-4k^2 - 20 = 0 ?
campbell_st
  • campbell_st
so you have \[[-(2k + 3)]^2 - 4\times (k + 5) \times (k + 1)] = 0\]
anonymous
  • anonymous
4k^2+9+12k-4k^2 - 20 = 0 ?
campbell_st
  • campbell_st
well the 1st part is ok... b^2 = 4k^2 + 12k + 9 4ac = 4(k^2 + 6k + 5) then b^2 - 4ac : 4k^2 + 12k + 9 - 4k^2 - 24k -20 = 0 so collect like terms and solve for k
anonymous
  • anonymous
4k^2 - 24k -20????? How???
campbell_st
  • campbell_st
ac: (k + 5)(k + 1) = k^2 + 6k + 5 4ac = 4k^2 + 24k + 20 does that make sense..?
anonymous
  • anonymous
No
campbell_st
  • campbell_st
so in the equation are you happy that a = (k + 5) and c = (k + 1)...?
campbell_st
  • campbell_st
so if you multiply a and c (k + 5)(k + 1) = k^2 + 6k + 5 does that make sense..?
anonymous
  • anonymous
Yes
campbell_st
  • campbell_st
so then 4ac = 4k^2 + 24k + 20 so b^2 - 4ac is 4k^2 + 12k + 9 - 4k^2 -24k - 20 = 0 collect like terms -12k - 11 = 0
anonymous
  • anonymous
Now?
campbell_st
  • campbell_st
solve for k... -12k = 11 etc
anonymous
  • anonymous
k = 11/12
anonymous
  • anonymous
0.92?
campbell_st
  • campbell_st
well if -12k = 11 k = - 11/12 I'd leave it as a fraction.
anonymous
  • anonymous
- 11/12?? It should be +???
campbell_st
  • campbell_st
is that the answer you have been provided..?
anonymous
  • anonymous
I have no options.............
campbell_st
  • campbell_st
well are you happy that after collecting like terms -12k - 11 = 0 then you can multiply every term by -1 and get 12k + 11 = 0 and then you still get 12k = -11 so k = -11/12
anonymous
  • anonymous
Thanks.

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