- #1

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How integrate this task?

[itex]x^2+2x-5=0[/itex]

[itex]D=24[/itex], so I can't get real root.

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- Thread starter Tasy
- Start date

- #1

- 1

- 0

How integrate this task?

[itex]x^2+2x-5=0[/itex]

[itex]D=24[/itex], so I can't get real root.

- #2

Curious3141

Homework Helper

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How integrate this task?

[itex]x^2+2x-5=0[/itex]

[itex]D=24[/itex], so I can't get real root.

OK, that line with the discriminant made no sense. The discriminant is greater than zero, so clearly the quadratic does have real, distinct roots.

Hint: start off with completing the square on the denominator. In other words, express the denominator in the form [itex](x+a)^2 + c[/itex] and go from there.

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