What is 1+sqrt(7x-3)>3 ?

The solution to 1+sqrt(7x-3)>3 is x>1

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1+sqrt(7x-3)>3

1 + 7 x - 3 > 3

x > 1

Solution steps

1 + 7 x - 3 > 3

Move

1

to the right side

7 x - 3 > 2

Square both sides

(7 x - 3)^{2} > 2^{2}

Simplify

7 x - 3 > 4

7 x - 3 > 4 : x > 1

x > 1

Find singularity points

Find non-negative values for radicals:

x \geq \frac{3}{7}

Combine the intervals

x > 1 and x \geq \frac{3}{7}

Merge Overlapping Intervals

x > 1

6 + x = - 54

t^{2} + 7 t - 1 = 0

x^{2} - 27 x + 50 = 0

2 - 2

(x^{3})^{\frac{2}{3}} y^{3}