What is (4x-6)^2>= (6x-4)^2 ?

The solution to (4x-6)^2>= (6x-4)^2 is -1<= x<= 1

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(4x-6)^2>= (6x-4)^2

(4 x - 6)^{2} \geq (6 x - 4)^{2}

- 1 \leq x \leq 1

Interval Notation

[- 1, 1]

Solution steps

(4 x - 6)^{2} \geq (6 x - 4)^{2}

Rewrite in standard form

- x^{2} + 1 \geq 0

Move

1

to the right side

- x^{2} \geq - 1

Multiply both sides by

- 1

x^{2} \leq 1

For

u^{n} \leq a

, if

n

is even then

- 1 \leq x \leq 1

7 x^{9} - 4 > 17

x^{2} + x - 8 \geq x - x^{2} - 10

\frac{x + 1}{x ^{2} + 2} > \frac{1}{4}

\frac{( - x ^{2} + 4 x - 3 )}{( 2 x - 6 )} < 0

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