High School Equation: Commented Exercises and Competition Questions

One second degree equation is the whole equation in the form ax² + bx + c = 0, with a, b and c real numbers and a ≠ 0. To solve an equation of this type, you can use different methods.

Use the commented resolutions of the exercises below to clear all your doubts. Also be sure to test your knowledge with the resolved contest questions.

Commented Exercises

Exercise 1

My mom's age multiplied by my age equals 525. If when I was born my mother was 20 years old, how old am I?

Solution

Considering my age equal to x, we can then consider that my mother's age is equal to x + 20. How do we know the value of the product of our ages, then:

x. (x + 20) = 525

Applying to the distributive properties of multiplication:

x² + 20 x - 525 = 0

We then arrive at a complete 2nd degree equation, with a = 1, b = 20 and c = - 525.

To calculate the roots of the equation, that is, the values ​​of x where the equation equals zero, let's use Bhaskara's formula.

First, we must calculate the value of ∆:

To calculate the roots, we use:

Substituting the values ​​in the formula above, we will find the roots of the equation, like this:

As my age cannot be negative, we despise the value -35. So the result is 15 years.

Exercise 2

A square, represented in the figure below, has a rectangular shape and its area is equal to 1 350 m². Knowing that its width corresponds to 3/2 its height, determine the dimensions of the square.

Solution

Considering that its height is equal to x, the width will then be equal to 3/2x. The area of ​​a rectangle is calculated by multiplying its base by the height value. In this case, we have:

We arrive at an incomplete 2nd degree equation, with a = 3/2, b = 0 and c = - 1350, we can calculate this type of equation by isolating the x and calculating the square root value.

As the value of x represents the measure of height, we will disregard the - 30. Thus, the height of the rectangle is equal to 30 m. To calculate the width, let's multiply this value by 3/2:

Therefore, the square width is equal to 45 m and its height is equal to 30 m.

Exercise 3

So that x = 1 is the root of the equation 2ax² + (2nd² - a - 4) x - (2 + a²) = 0, the values ​​of a should be:

a) 3 and 2
b) - 1 and 1
c) 2 and - 3
d) 0 and 2
e) - 3 and - 2

Solution

To find the value of a, let's first replace x with 1. This way, the equation will look like this:

2.a.1² + (2nd² - to - 4). 1 - 2 - a² = 0
2nd + 2nd² - to - 4 - 2 - to² = 0
The² + to - 6 = 0

Now, we must calculate the root of the complete 2nd degree equation, for that we will use Bhaskara's formula.

Therefore, the correct alternative is the letter C.

Contest Questions

1) Epcar - 2017

Consider, in ℝ, the equation (m+2) x² - 2mx + (m - 1) = 0 in variable x, where m is a real number other than - 2.

Review the statements below and rate them as a V (TRUE) or F (FALSE).

( ) For all m > 2 the equation has an empty solution set.
( ) There are two real values ​​of m for the equation to admit equal roots.
( ) In the equation, if ∆ >0, then m can only assume positive values.

The correct sequence is

a) V - V - V
b) F - V - F
c) F - F - V
d) V - F - F

See Answer

Let's look at each of the statements:

For all m > 2 the equation has an empty solution set

Since the equation is of the second degree in ℝ, it will have no solution when the delta is less than zero. Calculating this value, we have:

So the first statement is true.

There are two real values ​​of m for the equation to admit of equal roots.

The equation will have equal real roots when Δ=0, that is:

- 4m + 8 =0
m=2

Therefore, the statement is false as there is only one value of m where the roots are real and equal.

In the equation, if ∆ >0, then m can only take positive values.

For Δ>0, we have:

Since there are in the set of infinite real numbers negative numbers less than 2, the statement is also false.

Alternative d: V-F-F

2) Coltec - UFMG - 2017

Laura has to solve a 2nd degree equation in the “home” but realizes that when copying from the blackboard to the notebook, she forgot to copy the coefficient of x. To solve the equation, he recorded it as follows: 4x² + ax + 9 = 0. Since she knew that the equation had only one solution, and this one was positive, she was able to determine the value of a, which is

a) – 13
b) – 12
c) 12
d) 13

See Answer

When an equation of the 2nd degree has a single solution, the delta, from Bhaskara's formula, is equal to zero. So to find the value of The, just calculate the delta, equaling its value to zero.

So if a = 12 or a = - 12 the equation will have only one root. However, we still need to check which of the values ​​of The the result will be a positive root.

For that, let's find the root, for the values ​​of The.

So for a = -12 the equation will have only one root and positive.

Alternative b: -12

3) Enem - 2016

A tunnel must be sealed with a concrete cover. The cross section of the tunnel and the concrete cover have the contours of a parabola arch and the same dimensions. To determine the cost of the work, an engineer must calculate the area under the parabolic arc in question. Using the horizontal axis at floor level and the symmetry axis of the parabola as the vertical axis, she obtained the following equation for the parabola:
y = 9 - x², where x and y are measured in meters.
It is known that the area under a parabola like this is equal to 2/3 of the area of ​​the rectangle whose dimensions are, respectively, equal to the base and height of the tunnel entrance.
What is the area of ​​the front of the concrete cover, in square metres?

a) 18
b) 20
c) 36
d) 45
e) 54

See Answer

To resolve this issue, we need to find the measurements of the base and height of the tunnel entrance, as the problem tells us that the area of ​​the front is equal to 2/3 of the area of ​​the rectangle with these dimensions.

These values ​​will be found from the 2nd degree equation given. The parabola of this equation has the concavity turned down, because the coefficient The is negative. Below is an outline of this parable.

From the graph, we can see that the measure of the base of the tunnel will be found by calculating the roots of the equation. Already its height, will be equal to the measure of the vertex.

To calculate the roots, we observe that the equation 9 - x² is incomplete, so we can find its roots by equating the equation to zero and isolating the x:

Therefore, the measurement of the base of the tunnel will be equal to 6 m, that is, the distance between the two roots (-3 and 3).

Looking at the graph, we see that the vertex point corresponds to the value on the y axis that x is equal to zero, so we have:

Now that we know the measurements of the tunnel's base and height, we can calculate its area:

Alternative c: 36

4) Cefet - RJ - 2014

For what value of "a" does the equation (x - 2).(2ax - 3) + (x - 2).(- ax + 1) = 0 have two roots and equal?

to 1
b) 0
c) 1
d) 2

See Answer

For a 2nd degree equation to have two equal roots, it is necessary that Δ=0, that is, b²-4ac=0. Before calculating the delta, we need to write the equation in the form ax² + bx + c = 0.

We can start by applying the distributive property. However, we note that (x - 2 ) is repeated in both terms, so let's put it in evidence:

(x - 2) (2ax -3 - ax + 1) = 0
(x - 2) (ax -2) = 0

Now, distributing the product, we have:

ax² - 2x - 2ax + 4 = 0

Calculating the Δ and equaling to zero, we find:

So when a = 1, the equation will have two equal roots.

Alternative c: 1

To learn more, see also:

• Second degree equation
• First Degree Equation
• Quadratic Function
• Quadratic Function - Exercises
• Linear Function
• Related Function Exercises