quadratic equations Archives - PH Civil Service Exam Reviewer

Solving Quadratic Word Problems in Algebra

Quadratic Equations are equations of the form ax^2 + bx + c = 0 where $a$ , $b$ and $c$ are real numbers and $a \neq 0$ . Depending on the form of the equation, you can solve for $x$ by extracting the quare root, factoring, or using the quadratic formula.This type of equation appears in various problems that involves multiplication and usually appears in the Civil Service Exams.

The following series details the method and strategies in solving problems involving quadratic equations.

How to Solve Quadratic Word Problems Part 1 is about solving problems involving consecutive integers. In this problem, the product of consecutive numbers is given and factoring was used to solve the problem. Continue Reading

How to Solve Quadratic Problems Part 2

By Civil Service Reviewer
on Jan 10, 2016
in Math Word Problems, Mathematics, Tutorials

In the previous post, we have used quadratic equations to solve a word problem involving consecutive numbers. In this post, we discuss more quadratic problems. This is the second problem in the series.

Problem 2

Miel is 12 years older than Nina. The product of their ages is 540.

Solution

Let x = age of Nina
x + 12 = age of Miel

The product of their ages is 540, so we can multiply the expressions above and equate the product to 540. That is,

x(x + 12) = 540.

Multiplying the expressions, we have

$x^2 + 12x = 540$ .

Subtracting 540 from both sides, we obtain

$x^2 + 12x - 540 = 0$ .

We want to find two numbers whose product is -540 and whose sum is 12. Those numbers are -18 and 30.

This means that the factors are

(x – 18)(x + 30) = 0.

Equating each expression to 0, we have

x – 18 = 0, x = 18
x + 30 = 0, x = – 30.

Since we are talking about age, we take the positive answer x = 18.

This means that Nina is 18 years old. Therefore, Miel is 18 + 12 = 30 years old.

Solving Quadratic Equations by Quadratic Formula

In the previous post, we have learned how to solve quadratic equations by factoring. In this post, we are going to learn how to solve quadratic equations using the quadratic formula. In doing this, we must identify the values of $a$ , $b$ , and $c$ , in $ax^2 + bx + c = 0$ and substitute their values to the quadratic formula

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac }}{2a}$ .

Note that the value of $a$ is the number in the term containing $x^2$ , $b$ is the number in the term containing $x$ , and $c$ is the value of the constant (without $x$ or $x^2$ ).

The results in this calculation which are the values of $x$ are the roots of the quadratic equation. Before you calculate using this formula, it is important that you master properties of radical numbers and how to calculate using them.

Example 1: Find the roots of $x^2 - 4x - 4 = 0$

Solution

From the equation, we can identify $a = 1$ , $b = -4$ , and $c = -4$ .

Substituting these values in the quadratic formula, we have

$x = \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-4)}}{2(1)}$
$x = \dfrac{4 \pm \sqrt{16 + 16}}{2}$
$x = \dfrac{4 \pm \sqrt{32}}{2}$ .

We know, that $\sqrt{32} = \sqrt{(16)(2)} = \sqrt{16} \sqrt{2} = 4 \sqrt{2}$ . So, we have

$x = \dfrac{4 \pm 4 \sqrt{2}}{2} = \dfrac{2(2 + \sqrt{2}}{2} = 2 \pm 2 \sqrt{2}$

Therefore, we have two roots

$2 + 2 \sqrt{2}$ or $2 - 2 \sqrt{2}$

Example 2: Find the roots of $2x^2 - 6x = 15$

Solution

Recall, that it easier to identify the values of $a$ , $b$ , and $c$ if the quadratic equation is in the general form which is $ax^2 + bx + c = 0$ . In order to make the right hand side of the equation above equal to 0, subtract 15 from both sides of the equation by 15. This results to

$2x^2 - 6x - 15$ .

As we can see, $a = 2$ , $b = -6$ and $c =-15$ .

Substituting these values to the quadratic formula, we have

$x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(2)(-15)}}{2(2)}$

$x = \dfrac{6 \pm \sqrt{36 + 120}}{4}$

$x = \dfrac{6 \pm \sqrt{156}}{4}$ .

But $\sqrt{156} = \sqrt{(4)(39)} = \sqrt{4} \sqrt{39} = 2 \sqrt{39}$ .

Therefore,

$x = \dfrac{6 \pm 2 \sqrt{39}}{4}$ .

Factoring out 2, we have

$\dfrac{2(3 \pm \sqrt{39}}{4} = \dfrac{3 \pm \sqrt{39}}{2}$

Therefore, we have two roots:

$\dfrac{3 + \sqrt{39}}{2}$ or $\dfrac{3 - \sqrt{39}}{2}$

That’s it. In the next post, we are going to learn how to use quadratic equations on how to solve word problems.

Solving Quadratic Equations by Factoring

By Civil Service Reviewer
on Dec 23, 2015
in Mathematics

In the previous post, we have learned how to solve quadratic equations by extracting the roots. In this post, we are going to learn how to solve quadratic equations by factoring.

To solve quadratic equations by factoring, we need to use the zero property of real numbers. It states that the product of two real numbers is zero if at least one of the two real numbers is zero. In effect, we need to transfer all the terms to the lefth hand side, let the right hand side be 0, equate factored form zero and find the value of x.

Example 1: $2x(x + 5) = 0$

Solution
$2x = 0, x = 0$
$x + 5 = 0, x = -5$ .

This means the solutions of $2x(x+5) =0$ are $0$ or $-5$ .

Example 2: $x^2 - x = 6$

Solution
Subtracting $6$ from both sides, we have
$x^2 - x - 6 = 0$ .

Factoring, we have
$(x - 3)(x + 2) = 0$
$x - 3 = 0, x = 3$
$x + 2 = 0, x = -2$ .

This means the solutions of $x^2 - x = 6$ are $3$ or $-2$ .

Example 3: $x^2 = x \sqrt{3}$

Solution

Subtracting $\sqrt{3}$ from both sides,
$x^2 - x \sqrt{3} = 0$ .

Factoring out $x$ , we have
$x(x - \sqrt{3}) = 0$ .

Equating both expression to 0, we have
$x = 0$
$x - \sqrt{3} = 0, x = \pm \sqrt{3}$ .

This means the solutions of $x^2 - x = 6$ are $0$ or $\pm \sqrt{3}$ .

Example 4: Solve $x^2 = 16$ .

Solution

$x^2 - 16 = 0$
$(x + 4)(x - 4) = 0$
$x = -4, x = 4$ .

This means the solutions of $x^2 = 16$ are $-4$ or $4$ .

Example 5: Solve $x^2 + 2x + 1 = 0$ .

Solution
$x^2 + 2x + 1 = 0$
$(x + 1)(x + 1) = 0$
$x + 1 = 0, x = -1$
$x + 1 = 0, x = -1$
$x = -1, x = -1$ .

This means the solutions of $x^2 + 2x + 1 = 0$ are $katex -1$ or $-1$ .

Example 6: Solve $(x - 3)^2 = 4x$

Solution
$(x - 3)^2 = 4x$
$x^2 - 6x + 9 = 4x$
$x^2 - 6x - 4x + 9 = 0$
$x^2 - 10x + 9 = 0$
$(x - 1)(x - 9) = 0$
$x = 1, x = 9$ .

This means the solutions of $(x - 3)^2 = 4x$ are $1$ or $9$ .

Example 7: $\frac{2x - 3}{x + 3} = \frac{x + 3}{x - 3}$
Solution

Cross multiplying, we have
$(2x - 3)(x - 3) = (x + 3)(x + 3)$ .

Expanding, we have
$2x^2 - 9x + 9 = x^2 + 6x + 9$ .

Transposing all the terms to the left hand side, we have
$x^2 - 15x = 0$
$x(x - 15) = 0$
$x = 0$
$x - 15 = 0$
$x = 15$ .

This means the solutions are $0$ or $15$ .

Solving Quadratic Equations by Extracting the Square Root

By Civil Service Reviewer
on Dec 22, 2015
in Mathematics

In the previous post, we have learned about quadratic equations or equations of the form $ax^2 + bx + c = 0$ , where a is not equal to 0. In this equation, we want to find the value of x which we call the root or the solution to the equation.

There are three strategies in finding the root of the equation: by extracting the roots, by completing the square, and by the quadratic formula. In this example, we will discuss, how to find the root of the quadratic equation by extracting the root.

Just like in solving equations, if we want to find the value of x, we put all the numbers on one side, and all the x’s on one side. Since quadratic equations contain the term $x^2$ , we can find the value of x by extracting the square roots. Below are five examples on how to do this.

Example 1: $2x^2 = 0$

Solution

Dividing both sides by 2, we have

$\frac{2x^2}{2} = \frac{0}{2}$ .

This gives us

$x^2 = 0$ .

Extracting the square root of both sides, we have

$x = 0$ .

Therefore, the root $x = 0$ .

Example 2: $x^2 - 36 = 0$

Solution

$x^2 - 36 = 0$ .

Adding 36 to both side, we have

$x^2 - 36 + 36 = 0 + 36$ .

$x^2 = 36$ .

Extracting the square root of both sides, we have

$\sqrt{x^2} = \sqrt{36}$ .

$x = \pm 6$ .

In this example, x has two roots: x = 6 and x = -6.

Example 3: $x^2 + 81 = 0$

Solution

Subtracting 81 from both sides, we have

$x^2 + 81 - 81 = 0 - 81$

$x^2 = -81$

$\sqrt{x^2} = - 81$

$x = \sqrt{-81}$ .

In this case, there is no number that when multiplied by itself is negative. For example, negative times negative is equal to positive, and positive times positive is equal to positive. Therefore, there is no real root. There is, however, what we call a complex root as shown in the video below.

Example 4: $5x^2 = 12$

Solution

$5x^2 = 12$

Dividing both sides by 5, we have

$x^2 = \frac{12}{5}$ .

Extracting the square root of both sides, we have

$\sqrt{x^2} = \sqrt{\frac{12}{5}} = \frac{\sqrt{12}}{\sqrt{5}}$

$x = \frac{\sqrt{4}\sqrt{3}}{\sqrt{5}}$

$x = \frac{2 \sqrt{3}}{\sqrt{5}}$ .

Example 5: $3x^2 = - 4$

$3x^2 = -4$

Dividing both sides by 3, we have

$x^2 = \frac{-4}{3}$ .

Extracting the square root of both sides, we have

$\sqrt{x^2} = \sqrt{\frac{-4}{3}}$ .

Again, the sign of the number inside the radical is negative, so there is no real root. To know how to compute for the complex root, watch the video below.

Browse Tag: quadratic equations

Solving Quadratic Word Problems in Algebra

How to Solve Quadratic Problems Part 2

Solving Quadratic Equations by Quadratic Formula

Solving Quadratic Equations by Factoring

Solving Quadratic Equations by Extracting the Square Root

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