What is 3x^2-4(x+1)<(2x-1)(x+1) ?

The solution to 3x^2-4(x+1)<(2x-1)(x+1) is (-sqrt(37)+5)/2 <x<(sqrt(37)+5)/2

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3x^2-4(x+1)<(2x-1)(x+1)

3 x^{2} - 4 (x + 1) < (2 x - 1) (x + 1)

\frac{- 37 + 5}{2} < x < \frac{37 + 5}{2}

Interval Notation

(\frac{- 37 + 5}{2}, \frac{37 + 5}{2})

Decimal Notation

- 0.54138 \dots < x < 5.54138 \dots

Solution steps

3 x^{2} - 4 (x + 1) < (2 x - 1) (x + 1)

Rewrite in standard form

x^{2} - 5 x - 3 < 0

Complete the square

x^{2} - 5 x - 3 : (x - \frac{5}{2})^{2} - \frac{37}{4}

(x - \frac{5}{2})^{2} - \frac{37}{4} < 0

Move

\frac{37}{4}

to the right side

(x - \frac{5}{2})^{2} < \frac{37}{4}

For

u^{n} < a

, if

n

is even then

- \frac{37}{4} < x - \frac{5}{2} < \frac{37}{4}

a < u < b

then

a < u and u < b

- \frac{37}{4} < x - \frac{5}{2} and x - \frac{5}{2} < \frac{37}{4}

- \frac{37}{4} < x - \frac{5}{2} : x > \frac{- 37 + 5}{2}

x - \frac{5}{2} < \frac{37}{4} : x < \frac{37 + 5}{2}

Combine the intervals

x > \frac{- 37 + 5}{2} and x < \frac{37 + 5}{2}

Merge Overlapping Intervals

\frac{- 37 + 5}{2} < x < \frac{37 + 5}{2}

x^{2} > \frac{4}{13}

\frac{( ( x + 5 ) ( x - 8 ) )}{( x - 12 )} > 0

7 x + 5 x < x^{2} - 10

\frac{( x - 2 ) ^{2} ( x - 1 )}{( x - 3 )} \geq 0

\frac{1}{( n - 2 )} + \frac{1}{( n - 1 )} > \frac{1}{n}

What is 3x^2-4(x+1)<(2x-1)(x+1) ?
The solution to 3x^2-4(x+1)<(2x-1)(x+1) is (-sqrt(37)+5)/2 <x<(sqrt(37)+5)/2