What is x^2-3x<2x^2-5x+6 ?

The solution to x^2-3x<2x^2-5x+6 is True for all x

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x^2-3x<2x^2-5x+6

x^{2} - 3 x < 2 x^{2} - 5 x + 6

True for all x

Interval Notation

(- \infty, \infty)

Solution steps

x^{2} - 3 x < 2 x^{2} - 5 x + 6

Rewrite in standard form

- x^{2} + 2 x - 6 < 0

Complete the square

- x^{2} + 2 x - 6 : - (x - 1)^{2} - 5

- (x - 1)^{2} - 5 < 0

Move

5

to the right side

- (x - 1)^{2} < 5

Multiply both sides by

- 1

(x - 1)^{2} > - 5

If n is even,

u^{n} \geq 0

for all

u

True for all x

x^{3} - x^{2} - 9 x + 9

6 x + 7 = 3 x - 6

\frac{2 x - 20}{3} = 2 x

515

a^{2} + 2 ab + b^{2}