What is 2x^2+5x-3>= x^2+1 ?

The solution to 2x^2+5x-3>= x^2+1 is x = (sqrt(41)-5)/2

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2x^2+5x-3>= x^2+1

2 x^{2} + 5 x - 3 \geq x^{2} + 1

x \leq \frac{- 41 - 5}{2} or x \geq \frac{41 - 5}{2}

Interval Notation

(- \infty, \frac{- 41 - 5}{2}] \cup [\frac{41 - 5}{2}, \infty)

Decimal Notation

x \leq - 5.70156 \dots or x \geq 0.70156 \dots

Solution steps

2 x^{2} + 5 x - 3 \geq x^{2} + 1

Rewrite in standard form

x^{2} + 5 x - 4 \geq 0

Complete the square

x^{2} + 5 x - 4 : (x + \frac{5}{2})^{2} - \frac{41}{4}

(x + \frac{5}{2})^{2} - \frac{41}{4} \geq 0

Move

\frac{41}{4}

to the right side

(x + \frac{5}{2})^{2} \geq \frac{41}{4}

For

u^{n} \geq a

, if

n

is even then

x + \frac{5}{2} \leq - \frac{41}{4} or x + \frac{5}{2} \geq \frac{41}{4}

x + \frac{5}{2} \leq - \frac{41}{4} : x \leq \frac{- 41 - 5}{2}

x + \frac{5}{2} \geq \frac{41}{4} : x \geq \frac{41 - 5}{2}

Combine the intervals

x \leq \frac{- 41 - 5}{2} or x \geq \frac{41 - 5}{2}

\frac{2 ^{- 0.1} - 1}{- 0.1}

\frac{( 10 ab ) ^{2} ( 2 a ^{4} b ^{3} )}{4 a ^{5} b}

(x + 5)^{2} - 64 = 0

f = (1)^{2} - 5

x^{2} - 5 x + 4

What is 2x^2+5x-3>= x^2+1 ?
The solution to 2x^2+5x-3>= x^2+1 is x<= (-sqrt(41)-5)/2 \lor x>= (sqrt(41)-5)/2