What is 4+5p-p^2>= 0 ?

The solution to 4+5p-p^2>= 0 is (-sqrt(41)+5)/2 <= p<= (sqrt(41)+5)/2

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4+5p-p^2>= 0

4 + 5 p - p^{2} \geq 0

\frac{- 41 + 5}{2} \leq p \leq \frac{41 + 5}{2}

Solution steps

4 + 5 p - p^{2} \geq 0

Complete the square

4 + 5 p - p^{2} : - (p - \frac{5}{2})^{2} + \frac{41}{4}

- (p - \frac{5}{2})^{2} + \frac{41}{4} \geq 0

Move

\frac{41}{4}

to the right side

- (p - \frac{5}{2})^{2} \geq - \frac{41}{4}

Multiply both sides by

- 1

(p - \frac{5}{2})^{2} \leq \frac{41}{4}

For

u^{n} \leq a

, if

n

is even then

- \frac{41}{4} \leq p - \frac{5}{2} \leq \frac{41}{4}

a \leq u \leq b

then

a \leq u and u \leq b

- \frac{41}{4} \leq p - \frac{5}{2} and p - \frac{5}{2} \leq \frac{41}{4}

- \frac{41}{4} \leq p - \frac{5}{2} : p \geq \frac{- 41 + 5}{2}

p - \frac{5}{2} \leq \frac{41}{4} : p \leq \frac{41 + 5}{2}

Combine the intervals

p \geq \frac{- 41 + 5}{2} and p \leq \frac{41 + 5}{2}

Merge Overlapping Intervals

\frac{- 41 + 5}{2} \leq p \leq \frac{41 + 5}{2}

2.7 \cdot 1 0^{- 4} \cdot m \leq 1.62

x^{2} + 12 x + 35 > 0

- 9 + 10

8 x^{2} - 2 x - 10

2 X^{2} + 3 X - 4 = 0

What is 4+5p-p^2>= 0 ?
The solution to 4+5p-p^2>= 0 is (-sqrt(41)+5)/2 <= p<= (sqrt(41)+5)/2