## anonymous one year ago (k+5)x^2 - (2k + 3)x + (k + 1) = 0 The roots are real and equal

1. anonymous

@freckles

2. campbell_st

real and equal means the discriminant is zero and the discriminant is found using $\Delta = b^2 - 4ac$

3. 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

4. anonymous

4k^2+9+12k-4k^2 - 20 = 0 ?

5. campbell_st

so you have $[-(2k + 3)]^2 - 4\times (k + 5) \times (k + 1)] = 0$

6. anonymous

4k^2+9+12k-4k^2 - 20 = 0 ?

7. 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

8. anonymous

4k^2 - 24k -20????? How???

9. campbell_st

ac: (k + 5)(k + 1) = k^2 + 6k + 5 4ac = 4k^2 + 24k + 20 does that make sense..?

10. anonymous

No

11. campbell_st

so in the equation are you happy that a = (k + 5) and c = (k + 1)...?

12. campbell_st

so if you multiply a and c (k + 5)(k + 1) = k^2 + 6k + 5 does that make sense..?

13. anonymous

Yes

14. 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

15. anonymous

Now?

16. campbell_st

solve for k... -12k = 11 etc

17. anonymous

k = 11/12

18. anonymous

0.92?

19. campbell_st

well if -12k = 11 k = - 11/12 I'd leave it as a fraction.

20. anonymous

- 11/12?? It should be +???

21. campbell_st

is that the answer you have been provided..?

22. anonymous

I have no options.............

23. 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

24. anonymous

Thanks.