2^2+5x<-x^2+1 — step-by-step solution

1) Let's solve in

ℝ

2^{2} + 5 x < - x^{2} + 1

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

2^{2}

2 \cdot 2

4

4 + 5 x < - x^{2} + 1

4

+ 5 x

+ x^{2}

- 1 < 0

ⓘ Start from

4

and move back

1

on the number line:

4 - 1

3

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

Given the quadratic form

a x^{2} + b x + c

, you can calculate the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = 5

and

c = 3

Δ = 5^{2} - 4 \cdot 1 \cdot 3

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

5^{2}

5 \cdot 5

25

Δ = 25 - 4 \cdot 3

ⓘ Multiplications are done before additions and subtractions:

4 \cdot 3

12

Δ = 25 - 12

ⓘ Start from

25

and move back

12

on the number line:

25 - 12

13

Δ = 13

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 5 - \sqrt{13}}{2 \cdot 1}

x_{1} = \frac{- 5 - \sqrt{13}}{2}

ⓘ The minus sign is factored out, changing the sign of each term:

- 5 - \sqrt{13}

- (5 + \sqrt{13})

x_{1} = - \frac{5 + \sqrt{13}}{2}

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 5 + \sqrt{13}}{2 \cdot 1}

x_{2} = \frac{- 5 + \sqrt{13}}{2}

The quadratic has the opposite sign to a =

1

between its roots, so it is indeed negative between its roots; the roots, where it is zero, are excluded.

▶Solution obtained:

x \in (- \frac{5 + \sqrt{13}}{2}, \frac{\sqrt{13} - 5}{2})

2) The numeric value of:

x \in (- \frac{5 + \sqrt{13}}{2}, \frac{\sqrt{13} - 5}{2})

equals:

x \in (- \frac{5 + \approx 3.6056}{2}, \frac{\approx 3.6056 - 5}{2})

x \in (- \frac{\approx 8.6056}{2}, - \frac{\approx 1.3944}{2})

▷

x \in (\approx -4.3028, \approx -0.6972)