The solution to x^2+8x-7<0 is -sqrt(23)-4<x<sqrt(23)-4

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x^2+8x-7<0

Solution

x^{2} + 8 x - 7 < 0

- 23 - 4 < x < 23 - 4

Interval Notation

(- 23 - 4, 23 - 4)

Decimal Notation

- 8.79583 \dots < x < 0.79583 \dots

Solution steps

x^{2} + 8 x - 7 < 0

Complete the square

x^{2} + 8 x - 7 : (x + 4)^{2} - 23

(x + 4)^{2} - 23 < 0

Move

23

to the right side

(x + 4)^{2} < 23

For

u^{n} < a

, if

n

is even then

- 23 < x + 4 < 23

a < u < b

then

a < u and u < b

- 23 < x + 4 and x + 4 < 23

- 23 < x + 4 : x > - 23 - 4

x + 4 < 23 : x < 23 - 4

Combine the intervals

x > - 23 - 4 and x < 23 - 4

Merge Overlapping Intervals

- 23 - 4 < x < 23 - 4

lo g_{\frac{2}{3}} (\frac{9}{4}) = x

(4 x - 1)^{2} = 11

0.373506 = \frac{x ^{2}}{42.1889 \cdot 0.2727}

16 x^{3} - 54 y^{3}

5 (n - 3) = 20