The solution to 7x^2+x-1<3 is (-sqrt(113)-1)/(14)<x<(sqrt(113)-1)/(14)

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7x^2+x-1<3

Solution

7 x^{2} + x - 1 < 3

\frac{- 113 - 1}{14} < x < \frac{113 - 1}{14}

Interval Notation

(\frac{- 113 - 1}{14}, \frac{113 - 1}{14})

Decimal Notation

- 0.83072 \dots < x < 0.68786 \dots

Solution steps

7 x^{2} + x - 1 < 3

Rewrite in standard form

7 x^{2} + x - 4 < 0

Complete the square

7 x^{2} + x - 4 : 7 (x + \frac{1}{14})^{2} - \frac{113}{28}

7 (x + \frac{1}{14})^{2} - \frac{113}{28} < 0

Move

\frac{113}{28}

to the right side

7 (x + \frac{1}{14})^{2} < \frac{113}{28}

Divide both sides by

7

(x + \frac{1}{14})^{2} < \frac{113}{196}

For

u^{n} < a

, if

n

is even then

- \frac{113}{196} < x + \frac{1}{14} < \frac{113}{196}

a < u < b

then

a < u and u < b

- \frac{113}{196} < x + \frac{1}{14} and x + \frac{1}{14} < \frac{113}{196}

- \frac{113}{196} < x + \frac{1}{14} : x > \frac{- 113 - 1}{14}

x + \frac{1}{14} < \frac{113}{196} : x < \frac{113 - 1}{14}

Combine the intervals

x > \frac{- 113 - 1}{14} and x < \frac{113 - 1}{14}

Merge Overlapping Intervals

\frac{- 113 - 1}{14} < x < \frac{113 - 1}{14}

3 x^{2} - 1 x - 10

4 x + 2 > - 10

2 x^{2} - 10 x - 6

6 x + 9 x^{3}

- 4 < 3 x + 2 < 5

What is 7x^2+x-1<3 ?
The solution to 7x^2+x-1<3 is (-sqrt(113)-1)/(14)<x<(sqrt(113)-1)/(14)