d(10d-9)+1=0 — step-by-step solution

1) Let's solve in

ℝ

d \cdot (10 d - 9) + 1 = 0

ⓘ Expand by multiplying each term inside the parentheses:

d \cdot (10 d - 9)

10 d^{2} + d \cdot (-9)

10 d^{2} + d \cdot (-9) + 1 = 0

ⓘ The minus sign is placed in front of the product:

d \cdot (-9)

- 9 d

10 d^{2} - 9 d + 1 = 0

Given the quadratic form

a d^{2} + b d + c

, you can calculate the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 10

b = -9

and

c = 1

Δ = {(-9)}^{2} - 4 \cdot 10 \cdot 1

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

{(-9)}^{2}

(-9) \cdot (-9)

81

Δ = 81 - 4 \cdot 10

ⓘ Multiplications are done before additions and subtractions:

4 \cdot 10

40

Δ = 81 - 40

ⓘ Start from

81

and move back

40

on the number line:

81 - 40

41

Δ = 41

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

d_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-9) - \sqrt{41}}{2 \cdot 10}

d_{1} = \frac{9 - \sqrt{41}}{2 \cdot 10}

ⅰ

2 \cdot 10

20

d_{1} = \frac{9 - \sqrt{41}}{20}

The quadratic formula gives

d_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-9) + \sqrt{41}}{2 \cdot 10}

d_{2} = \frac{9 + \sqrt{41}}{2 \cdot 10}

ⅰ

2 \cdot 10

20

d_{2} = \frac{9 + \sqrt{41}}{20}

▶Solutions obtained:

d = {\frac{9 - \sqrt{41}}{20}, \frac{9 + \sqrt{41}}{20}}

2) The numeric value of:

d = {\frac{9 - \sqrt{41}}{20}, \frac{9 + \sqrt{41}}{20}}

equals:

d = {\approx \frac{9 - 6.4031}{20}, \approx \frac{9 + 6.4031}{20}}

d = {\approx \frac{2.5969}{20}, \approx \frac{15.4031}{20}}

▷

d = {\approx 0.1298, \approx 0.7702}