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3 years ago

Solve the equation: 2x^2 + 3x = 2

Mathematics

2 answers:

ASHA 777 [7]3 years ago

8 0

Answer:

x = - 2, x = $\frac{1}{2}$

Step-by-step explanation:

Given

2x² + 3x = 2 ( subtract 2 from both sides )

2x² + 3x - 2 = 0

Consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term

product = 2 × - 2 = - 4 and sum = + 3

The factors are + 4 and - 1

Use these factors to split the x- term

2x² + 4x - x - 2 = 0 ( factor the first/second and third/fourth terms )

2x(x + 2) - 1(x + 2) = 0 ← factor out (x + 2) from each term

(x + 2)(2x - 1) = 0

Equate each factor to zero and solve for x

x + 2 = 0 ⇒ x = - 2

2x - 1 = 0 ⇒ 2x = 1 ⇒ x = $\frac{1}{2}$

Dafna11 [192]3 years ago

5 0

Answer:

x = -2 or x = 1/2

Step-by-step explanation:

First get the eq in standard form, which means:

x² should not have that 2 in front of it

move the other 2 to the left

so first let's divide by 2:

x² + 3/2 x = 1

now move the 1 to the left

x² + 3/2 x - 1 = 0

now look for two numbers that, when multiplied, give -1, and when added, give 3/2. This is a bit of trial-and-error if you want to solve in this way.

The numbers are 2 and -1/2, since 2*-1/2 = -1, and 2 - 1/2 = 3/2

The factorization thus is:

x² + 3/2 x - 1 = (x+2)(x-1/2) = 0

Either factor (x+2) or (x-1/2) can be zero to solve the solution, hence x = -2 or x = 1/2

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Find the values of the sine, cosine, and tangent for ZA C A 36ft B 24ft

Reptile [31]

<h2>Question:</h2>

Find the values of the sine, cosine, and tangent for ∠A

a. sin A = $\frac{\sqrt{13} }{2}$ , cos A = $\frac{\sqrt{13} }{3}$ , tan A = $\frac{2 }{3}$

b. sin A = $3\frac{\sqrt{13} }{13}$ , cos A = $2\frac{\sqrt{13} }{13}$ , tan A = $\frac{3}{2}$

c. sin A = $\frac{\sqrt{13} }{3}$ , cos A = $\frac{\sqrt{13} }{2}$ , tan A = $\frac{3}{2}$

d. sin A = $2\frac{\sqrt{13} }{13}$ , cos A = $3\frac{\sqrt{13} }{13}$ , tan A = $\frac{2 }{3}$

<h2>Answer:</h2>

d. sin A = $2\frac{\sqrt{13} }{13}$ , cos A = $3\frac{\sqrt{13} }{13}$ , tan A = $\frac{2 }{3}$

<h2>Step-by-step explanation:</h2>

The triangle for the question has been attached to this response.

As shown in the triangle;

AC = 36ft

BC = 24ft

ACB = 90°

To calculate the values of the sine, cosine, and tangent of ∠A;

i. First calculate the value of the missing side AB.

Using Pythagoras' theorem;

⇒ (AB)² = (AC)² + (BC)²

Substitute the values of AC and BC

⇒ (AB)² = (36)² + (24)²

Solve for AB

⇒ (AB)² = 1296 + 576

⇒ (AB)² = 1872

⇒ AB = $\sqrt{1872}$

⇒ AB = $12\sqrt{13}$ ft

From the values of the sides, it can be noted that the side AB is the hypotenuse of the triangle since that is the longest side with a value of $12\sqrt{13}$ ft (43.27ft).

ii. Calculate the sine of ∠A (i.e sin A)

The sine of an angle (Ф) in a triangle is given by the ratio of the opposite side to that angle to the hypotenuse side of the triangle. i.e

sin Ф = $\frac{opposite}{hypotenuse}$ -------------(i)

In this case,

Ф = A

opposite = 24ft (This is the opposite side to angle A)

hypotenuse = $12\sqrt{13}$ ft (This is the longest side of the triangle)

Substitute these values into equation (i) as follows;

sin A = $\frac{24}{12\sqrt{13} }$

sin A = $\frac{2}{\sqrt{13}}$

Rationalize the result by multiplying both the numerator and denominator by $\sqrt{13}$

sin A = $\frac{2}{\sqrt{13}} * \frac{\sqrt{13} }{\sqrt{13} }$

sin A = $\frac{2\sqrt{13} }{13}$

iii. Calculate the cosine of ∠A (i.e cos A)

The cosine of an angle (Ф) in a triangle is given by the ratio of the adjacent side to that angle to the hypotenuse side of the triangle. i.e

cos Ф = $\frac{adjacent}{hypotenuse}$ -------------(ii)

In this case,

Ф = A

adjacent = 36ft (This is the adjecent side to angle A)

hypotenuse = $12\sqrt{13}$ ft (This is the longest side of the triangle)

Substitute these values into equation (ii) as follows;

cos A = $\frac{36}{12\sqrt{13} }$

cos A = $\frac{3}{\sqrt{13}}$

Rationalize the result by multiplying both the numerator and denominator by $\sqrt{13}$

cos A = $\frac{3}{\sqrt{13}} * \frac{\sqrt{13} }{\sqrt{13} }$

cos A = $\frac{3\sqrt{13} }{13}$

iii. Calculate the tangent of ∠A (i.e tan A)

The cosine of an angle (Ф) in a triangle is given by the ratio of the opposite side to that angle to the adjacent side of the triangle. i.e

tan Ф = $\frac{opposite}{adjacent}$ -------------(iii)

In this case,

Ф = A

opposite = 24 ft (This is the opposite side to angle A)

adjacent = 36 ft (This is the adjacent side to angle A)

Substitute these values into equation (iii) as follows;

tan A = $\frac{24}{36}$

tan A = $\frac{2}{3}$

6 0

3 years ago