The solution to 5x+x^2>100 is x (5sqrt(17)-5)/2

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5x+x^2>100

Solution

5 x + x^{2} > 100

x < \frac{- 5 17 - 5}{2} or x > \frac{5 17 - 5}{2}

Interval Notation

(- \infty, \frac{- 5 17 - 5}{2}) \cup (\frac{5 17 - 5}{2}, \infty)

Decimal Notation

x < - 12.80776 \dots or x > 7.80776 \dots

Solution steps

5 x + x^{2} > 100

Rewrite in standard form

5 x + x^{2} - 100 > 0

Complete the square

5 x + x^{2} - 100 : (x + \frac{5}{2})^{2} - \frac{425}{4}

(x + \frac{5}{2})^{2} - \frac{425}{4} > 0

Move

\frac{425}{4}

to the right side

(x + \frac{5}{2})^{2} > \frac{425}{4}

For

u^{n} > a

, if

n

is even then

x + \frac{5}{2} < - \frac{425}{4} or x + \frac{5}{2} > \frac{425}{4}

x + \frac{5}{2} < - \frac{425}{4} : x < \frac{- 5 17 - 5}{2}

x + \frac{5}{2} > \frac{425}{4} : x > \frac{5 17 - 5}{2}

Combine the intervals

x < \frac{- 5 17 - 5}{2} or x > \frac{5 17 - 5}{2}

\frac{3}{4} - 2

∣ x ∣ + 3 < 5

lo g_{15} (1575) - lo g_{15} (7)

\frac{1}{8} + \frac{1}{2} x = \frac{1}{4}

1.754 \cdot 1 0^{- 5} = \frac{x ^{2}}{0.1}

What is 5x+x^2>100 ?
The solution to 5x+x^2>100 is x<(-5sqrt(17)-5)/2 \lor x>(5sqrt(17)-5)/2