What is sqrt(-x^2+8x-7)<x-1 ?

The solution to sqrt(-x^2+8x-7)<x-1 is 4<x<= 7

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sqrt(-x^2+8x-7)<x-1

- x^{2} + 8 x - 7 < x - 1

4 < x \leq 7

Interval Notation

(4, 7]

Solution steps

- x^{2} + 8 x - 7 < x - 1

f (x) < g (x)

then

g (x) > 0

x - 1 > 0 : x > 1

- x^{2} + 8 x - 7 < x - 1 : x < 1 or x > 4

Find singularity points

Find non-negative values for radicals:

1 \leq x \leq 7

Combine the intervals

x > 1 and (x < 1 or x > 4) and 1 \leq x \leq 7

Merge Overlapping Intervals

4 < x \leq 7

Verify Solutions:

4 < x \leq 7

True

The solution is

4 < x \leq 7

\frac{( ∣ x + 2 ∣ - ∣ x ∣ )}{( 8 - x ^{2} )} \geq 0

\frac{( x - 3 )}{( x + 1 )} \geq \frac{( x + 2 )}{( x - 4 )}

x - 3 < (15 - x)

x - 2 < x - 4

x^{2} - 3 x + 2 > 2 x - 5