What is 7x+3>= (x^2+2) ?

The solution to 7x+3>= (x^2+2) is (-sqrt(53)+7)/2 <= x<= (sqrt(53)+7)/2

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7x+3>= (x^2+2)

7 x + 3 \geq (x^{2} + 2)

\frac{- 53 + 7}{2} \leq x \leq \frac{53 + 7}{2}

Interval Notation

[\frac{- 53 + 7}{2}, \frac{53 + 7}{2}]

Decimal Notation

- 0.14005 \dots \leq x \leq 7.14005 \dots

Solution steps

7 x + 3 \geq x^{2} + 2

Rewrite in standard form

7 x + 1 - x^{2} \geq 0

Complete the square

7 x + 1 - x^{2} : - (x - \frac{7}{2})^{2} + \frac{53}{4}

- (x - \frac{7}{2})^{2} + \frac{53}{4} \geq 0

Move

\frac{53}{4}

to the right side

- (x - \frac{7}{2})^{2} \geq - \frac{53}{4}

Multiply both sides by

- 1

(x - \frac{7}{2})^{2} \leq \frac{53}{4}

For

u^{n} \leq a

, if

n

is even then

- \frac{53}{4} \leq x - \frac{7}{2} \leq \frac{53}{4}

a \leq u \leq b

then

a \leq u and u \leq b

- \frac{53}{4} \leq x - \frac{7}{2} and x - \frac{7}{2} \leq \frac{53}{4}

- \frac{53}{4} \leq x - \frac{7}{2} : x \geq \frac{- 53 + 7}{2}

x - \frac{7}{2} \leq \frac{53}{4} : x \leq \frac{53 + 7}{2}

Combine the intervals

x \geq \frac{- 53 + 7}{2} and x \leq \frac{53 + 7}{2}

Merge Overlapping Intervals

\frac{- 53 + 7}{2} \leq x \leq \frac{53 + 7}{2}

\frac{1}{3} + \frac{2}{3} - \frac{7}{3}

\frac{( \frac{3 x + y}{2 x y} )}{( \frac{5 x}{x + 2} )}

∣ 10 x + 2 ∣ > 5

3 x - 15 > 3

2 π \cdot 60

What is 7x+3>= (x^2+2) ?
The solution to 7x+3>= (x^2+2) is (-sqrt(53)+7)/2 <= x<= (sqrt(53)+7)/2