creampoof
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  • 13-08-2020
  • Mathematics
contestada

Find the sum of all $r$ such that $\frac{8r^2 - 14 r + 3}{r+5} = 4r -1$.

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FaDaReAdEoLa
FaDaReAdEoLa FaDaReAdEoLa
  • 13-08-2020

Answer:

8.25

Step-by-step explanation:

[tex]\frac{8r^2 - 14 r + 3}{r+5} = 4r -1[/tex]

[tex]8r^2 - 14r + 3= (4r -1)(r + 5)[/tex]

[tex]8r^2 - 14r + 3= 4 {r}^{2} + 19r - 5[/tex]

[tex]4 {r}^{2} - 33r + 8=0[/tex]

[tex]4 {r}^{2} - 32r - r + 8 = 0[/tex]

[tex]4r(r - 8) - 1(r - 8) = 0[/tex]

[tex](4r - 1)(r - 8)[/tex]

[tex]r = 8 \: or \: \frac{1}{4} [/tex]

Therefore the sum of all values of r

[tex]8 + \frac{1}{4} = 8 \frac{1}{4} [/tex]

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