7(x-4)^2-3(x+5)^2=4(x+1)-2 — step-by-step solution

1) Let's solve in

ℝ

7 \cdot {(x - 4)}^{2} - 3 \cdot {(x + 5)}^{2} = 4 \cdot (x + 1) - 2

7 \cdot {(x - 4)}^{2}

- 3 \cdot {(x + 5)}^{2}

- 4 \cdot (x + 1)

+ 2 = 0

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

{(x - 4)}^{2}

(x - 4) \cdot (x - 4)

{(x + 5)}^{2}

(x + 5) \cdot (x + 5)

■Expand by multiplying each term inside the parentheses:

- 4 \cdot (x + 1)

- (4 x + 4)

7 \cdot (x - 4) \cdot (x - 4)

- 3 \cdot (x + 5) \cdot (x + 5)

- (4 x + 4)

+ 2 = 0

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

- (4 x + 4)

- 4 x - 4

7 \cdot (x - 4) \cdot (x - 4)

- 3 \cdot (x + 5) \cdot (x + 5)

- 4 x

- 4

+ 2 = 0

ⓘ Start from

-4

and move forward

2

on the number line, getting closer to zero:

- 4 + 2

-2

7 \cdot (x - 4) \cdot (x - 4)

- 3 \cdot (x + 5) \cdot (x + 5)

- 4 x

- 2 = 0

ⓘ Expand by multiplying each term inside the parentheses:

7 \cdot (x - 4)

7 x - 28

3 \cdot (x + 5)

3 x + 15

(7 x - 28) \cdot (x - 4)

- (3 x + 15) \cdot (x + 5)

- 4 x

- 2 = 0

ⓘ Expand by multiplying each term inside the parentheses:

(7 x - 28) \cdot (x - 4)

7 \cdot x \cdot x

+ 7 x \cdot (-4)

- 28 x

- 28 \cdot (-4)

- (3 x + 15) \cdot (x + 5)

- (3 \cdot x \cdot x

+ 3 x \cdot 5

+ 15 x

+ 15 \cdot 5)

7 \cdot x \cdot x

+ 7 x \cdot (-4)

- 28 x

- 28 \cdot (-4)

- (3 \cdot x \cdot x

+ 3 x \cdot 5

+ 15 x

+ 15 \cdot 5)

- 4 x

- 2 = 0

ⓘ Multiplying powers with the same base adds the exponents:

x \cdot x

x^{2}

■Multiplications are done before additions and subtractions. A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

7 \cdot (-4)

-28

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

-28 \cdot (-4)

28 \cdot 4

(20 + 8) \cdot 4

20 \cdot 4 + 8 \cdot 4

80 + 32

112

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

- (3 \cdot x \cdot x

+ 3 x \cdot 5

+ 15 x

+ 15 \cdot 5)

- 3 \cdot x \cdot x

- 3 x \cdot 5

- 15 x

- 15 \cdot 5

■Like terms are grouped by adding their coefficients:

- 28 x - 4 x

-32 x

7 x^{2}

- 28 x

- 32 x

+ 112

- 3 \cdot x \cdot x

- 3 x \cdot 5

- 15 x

- 15 \cdot 5

- 2 = 0

ⓘ Multiplying powers with the same base adds the exponents:

x \cdot x

x^{2}

■Multiplications are done before additions and subtractions:

3 \cdot 5

15

15 \cdot 5

(10 + 5) \cdot 5

10 \cdot 5 + 5 \cdot 5

50 + 25

75

■Like terms are grouped by adding their coefficients:

- 28 x - 32 x - 15 x

-75 x

7 x^{2}

- 75 x

+ 112

- 3 x^{2}

- 15 x

- 75

- 2 = 0

ⓘ Start from

112

and move back

75

on the number line:

112 - 75

37

■Like terms are grouped by adding their coefficients:

7 x^{2} - 3 x^{2}

4 x^{2}

- 75 x - 15 x

-90 x

4 x^{2}

- 90 x

+ 37

- 2 = 0

ⓘ Start from

37

and move back

2

on the number line:

37 - 2

35

4 x^{2} - 90 x + 35 = 0

Given the quadratic form

a x^{2} + b x + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 4

b = -90

and

c = 35

Δ = {(-90)}^{2} - 4 \cdot 4 \cdot 35

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

{(-90)}^{2}

(-90) \cdot (-90)

8100

■Multiplications are done before additions and subtractions:

4 \cdot 4

16

Δ = 8100 - 16 \cdot 35

ⓘ Multiplications are done before additions and subtractions:

16 \cdot 35

(10 + 6) \cdot 35

10 \cdot 35 + 6 \cdot 35

350 + 210

560

Δ = 8100 - 560

ⓘ Start from

8100

and move back

560

on the number line:

8100 - 560

7540

Δ = 7540

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-90) - \sqrt{7540}}{2 \cdot 4}

x_{1} = \frac{90 - \sqrt{7540}}{2 \cdot 4}

ⓘ We look for the largest square dividing

7540

4

2^{2}

, and take it out of the root:

\sqrt{7540}

\sqrt{4 \cdot 1885}

\sqrt{4} \cdot \sqrt{1885}

2 \cdot \sqrt{1885}

ⅰ

2 \cdot 4

8

x_{1} = \frac{90 - 2 \cdot \sqrt{1885}}{8}

ⓘ A fraction is simplified by dividing the numerator and the denominator by the same number:

\frac{90 - 2 \cdot \sqrt{1885}}{8}

\frac{45 - \sqrt{1885}}{4}

x_{1} = \frac{45 - \sqrt{1885}}{4}

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-90) + \sqrt{7540}}{2 \cdot 4}

x_{2} = \frac{90 + \sqrt{7540}}{2 \cdot 4}

ⓘ We look for the largest square dividing

7540

4

2^{2}

, and take it out of the root:

\sqrt{7540}

\sqrt{4 \cdot 1885}

\sqrt{4} \cdot \sqrt{1885}

2 \cdot \sqrt{1885}

ⅰ

2 \cdot 4

8

x_{2} = \frac{90 + 2 \cdot \sqrt{1885}}{8}

ⓘ A fraction is simplified by dividing the numerator and the denominator by the same number:

\frac{90 + 2 \cdot \sqrt{1885}}{8}

\frac{45 + \sqrt{1885}}{4}

x_{2} = \frac{45 + \sqrt{1885}}{4}

▶Solutions found:

x = {\frac{45 - \sqrt{1885}}{4}, \frac{45 + \sqrt{1885}}{4}}

2) The numeric value of:

x = {\frac{45 - \sqrt{1885}}{4}, \frac{45 + \sqrt{1885}}{4}}

is:

x = {\approx \frac{45 - 43.4166}{4}, \approx \frac{45 + 43.4166}{4}}

x = {\approx \frac{1.5834}{4}, \approx \frac{88.4166}{4}}

▷

x = {\approx 0.3959, \approx 22.1041}