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3 years ago

Write the quadratic equation whose roots are 3 and -1, and whose leading coefficient is 5.

Mathematics

1 answer:

Assoli18 [71]3 years ago

6 0

Answer:

5x² - 10x - 15 = 0

Step-by-step explanation:

Given that the roots are x = 3 and x = - 1, then the factors are

(x - 3) and (x + 1) and the quadratic is the product of the factors, that is

f(x) = a(x - 3)(x + 1) ← a is a multiplier

Here a = 5, thus

f(x) = 5(x - 3)(x + 1) ← expand factors using FOIL

= 5(x² - 2x - 3) ← distribute parenthesis by 5

= 5x² - 10x - 15

Thus equation is

5x² - 10x - 15 = 0

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Consider a triangle ABC like the one below. Suppose that a = 31, b = 23, and c = 20. (The figure is not drawn to scale.) Solve t

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<h3>How to solve the triangle?</h3>

The figure is not given;

However, the question can still be solved without it

The given parameters are:

a = 31, b = 23, and c = 20

Calculate angle A using the following law of cosine

a² = b² + c² - 2bc * cos(A)

So, we have:

31² = 23² + 20² - 2 * 23 * 20 * cos(A)

Evaluate

961 = 929 - 920 * cos(A)

Subtract 929 from both sides

32 =- 920 * cos(A)

Divide both sides by -920

cos(A) = -0.0348

Take the arc cos of both sides

A = 92.0

Calculate angle B using the following law of sine

a/sin(A) = b/sin(B)

So, we have:

31/sin(92) = 23/sin(B)

This gives

31.0189 = 23/sin(B)

Rewrite as:

sin(B) =23/31.0189

Evaluate

sin(B) =0.7415

Take arc sin of both sides

B = 47.9

Calculate angle C using:

C = 180 - 92.0 - 47.9

Evaluate

C = 40.1

Hence, the solution to the triangle is A = 92.0, B = 47.9 and C = 40.1