2v^2+v-5=(v+1)^2 — step-by-step solution

Let's solve in

ℝ

2 v^{2} + v - 5 = {(v + 1)}^{2}

2 v^{2}

+ v

- 5

- {(v + 1)}^{2} = 0

ⓘ Raising a sum to an integer power is multiplying it by itself:

{(v + 1)}^{2}

(v + 1) \cdot (v + 1)

2 v^{2}

+ v

- 5

- (v + 1) \cdot (v + 1) = 0

ⓘ Expand by multiplying each term inside the parentheses:

- (v + 1) \cdot (v + 1)

- (v \cdot v

+ v \cdot 1

+ 1 \cdot v

+ 1 \cdot 1)

2 v^{2}

+ v

- 5

- (v \cdot v

+ v \cdot 1

+ 1 \cdot v

+ 1 \cdot 1) = 0

ⓘ A minus sign before a parenthesis changes the sign of each of its terms:

- (v \cdot v

+ v \cdot 1

+ 1 \cdot v

+ 1 \cdot 1)

- v \cdot v

- v \cdot 1

- 1 \cdot v

- 1 \cdot 1

2 v^{2}

+ v

- 5

- v \cdot v

- v \cdot 1

- 1 \cdot v

- 1 \cdot 1 = 0

ⓘ Multiplying powers with the same base adds the exponents:

v \cdot v

v^{2}

■Multiplying by 1 changes nothing:

1 \cdot v

v

2 v^{2}

+ v

- 5

- v^{2}

- v

- v

- 1 = 0

ⓘ Start from

-5

and move back

1

more on the number line, away from zero:

- 5 - 1

-6

■Like terms are grouped by adding their coefficients:

2 v^{2} - v^{2}

v^{2}

v

- 6

- v

- v

- v - 6

v^{2} - v - 6 = 0

Given the quadratic form

a v^{2} + b v + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = -1

and

c = -6

Δ = {(-1)}^{2} - 4 \cdot 1 \cdot (-6)

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

{(-1)}^{2}

(-1) \cdot (-1)

1

Δ = 1 - 4 \cdot (-6)

ⓘ Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive:

- 4 \cdot (-6)

24

Δ = 1 + 24

ⓘ Start from

1

and move forward

24

on the number line:

1 + 24

25

Δ = 25

Δ > 0

, the equation has two roots:

The quadratic formula gives

v_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-1) - \sqrt{25}}{2 \cdot 1}

v_{1} = \frac{1 - \sqrt{25}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{25}

5

;

5^{2}

25

v_{1} = \frac{1 - 5}{2}

ⓘ The numerator and the denominator are computed before the division:

1 - 5

-4

v_{1} = - \frac{4}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{4}{2}

-2

v_{1} = -2

The quadratic formula gives

v_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-1) + \sqrt{25}}{2 \cdot 1}

v_{2} = \frac{1 + \sqrt{25}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{25}

5

;

5^{2}

25

v_{2} = \frac{1 + 5}{2}

ⓘ The numerator and the denominator are computed before the division:

1 + 5

6

v_{2} = \frac{6}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{6}{2}

3

v_{2} = 3

▶Solutions obtained:

v = {-2, 3}