The solution to x(7-x)>8 is (-sqrt(17)+7)/2 <x<(sqrt(17)+7)/2

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x(7-x)>8

Solution

x (7 - x) > 8

\frac{- 17 + 7}{2} < x < \frac{17 + 7}{2}

Solution steps

x (7 - x) > 8

Rewrite in standard form

7 x - x^{2} - 8 > 0

Complete the square

7 x - x^{2} - 8 : - (x - \frac{7}{2})^{2} + \frac{17}{4}

- (x - \frac{7}{2})^{2} + \frac{17}{4} > 0

Move

\frac{17}{4}

to the right side

- (x - \frac{7}{2})^{2} > - \frac{17}{4}

Multiply both sides by

- 1

(x - \frac{7}{2})^{2} < \frac{17}{4}

For

u^{n} < a

, if

n

is even then

- \frac{17}{4} < x - \frac{7}{2} < \frac{17}{4}

a < u < b

then

a < u and u < b

- \frac{17}{4} < x - \frac{7}{2} and x - \frac{7}{2} < \frac{17}{4}

- \frac{17}{4} < x - \frac{7}{2} : x > \frac{- 17 + 7}{2}

x - \frac{7}{2} < \frac{17}{4} : x < \frac{17 + 7}{2}

Combine the intervals

x > \frac{- 17 + 7}{2} and x < \frac{17 + 7}{2}

Merge Overlapping Intervals

\frac{- 17 + 7}{2} < x < \frac{17 + 7}{2}

\frac{( y ^{2} - a )}{y + 1}

2 \cdot x

c - 3 > 1

2 x (x + 1) + 7 (x + 1)^{2}

lo g_{5} (1) = y

What is x(7-x)>8 ?
The solution to x(7-x)>8 is (-sqrt(17)+7)/2 <x<(sqrt(17)+7)/2