What is 0>= sqrt(7x^2-57x+56) ?

The solution to 0>= sqrt(7x^2-57x+56) is x= 8/7 \lor x=7

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0>= sqrt(7x^2-57x+56)

0 \geq 7 x^{2} - 57 x + 56

x = \frac{8}{7} or x = 7

Interval Notation

x = \frac{8}{7} \cup x = 7

Decimal Notation

x = 1.14285 \dots or x = 7

Solution steps

0 \geq 7 x^{2} - 57 x + 56

Switch sides

7 x^{2} - 57 x + 56 \leq 0

Square both sides

(7 x^{2} - 57 x + 56)^{2} \leq 0^{2}

Simplify

7 x^{2} - 57 x + 56 \leq 0

7 x^{2} - 57 x + 56 \leq 0 : \frac{8}{7} \leq x \leq 7

\frac{8}{7} \leq x \leq 7

Find singularity points

Find non-negative values for radicals:

x \leq \frac{8}{7} or x \geq 7

Combine the intervals

\frac{8}{7} \leq x \leq 7 and (x \leq \frac{8}{7} or x \geq 7)

Merge Overlapping Intervals

x = \frac{8}{7}, x = 7

x = \frac{8}{7} or x = 7

(1)^{2} + (3)^{2}

(- 5 c d) (- 3 c^{4} d) - (7 c^{2} d^{2}) (2 c^{3})

(81 y^{7})^{\frac{1}{3}}

- 3 (x - 2) + 2 < 2 x - 3 - 3 x

144 a = 0.5

What is 0>= sqrt(7x^2-57x+56) ?
The solution to 0>= sqrt(7x^2-57x+56) is x= 8/7 \lor x=7