General solution of a quadratic equations

General Solution of Quadratic Equations

The Quadratic Formula

The quadratic formula used for solving quadratic equations is:

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

where $a$, $b$, and $c$ are the coefficients of the quadratic equation $ax^2 + bx + c = 0$.

Example 1

Solve $5x^2 - 8x + 3 = 0$ using the quadratic formula.

Example 2

Solve this quadratic equation using the quadratic formula: $3x^2 = -7x - 4$

Solution:

First, rewrite the equation in standard form:

$3x^2 + 7x + 4 = 0$

Here, $a = 3$, $b = 7$, $c = 4$

$x = \frac{-7 \pm \sqrt{7^2 - 4(3)(4)}}{2(3)}$

$x = \frac{-7 \pm \sqrt{49 - 48}}{6}$

$x = \frac{-7 \pm 1}{6}$

Therefore:

$x = \frac{-7 + 1}{6} = \frac{-6}{6} = -1$

or

$x = \frac{-7 - 1}{6} = \frac{-8}{6} = -\frac{4}{3}$

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Word Problems Leading to Quadratic Equations

Given a word problem, the following steps are used to recognize and solve the type of equation.

1. Choose the variables to represent the information.
2. Formulate the equation according to the information given.
3. Solve the equation using any of the methods you know.

In order to be sure with your answers, check if the solution you obtained is correct.

Example 3

The length of a rectangular plot is 8 centimeters more than the width. If the area of the plot is 240 cm², find the dimensions of length and width.

Solution

Let the width be $x$ cm.

The length of the plot is 8 more than the width, so the length is $x + 8$ cm.

We are given the area of the plot = 240 cm², and the area of a rectangle is given by length × width.

Then $(x + 8) \times x = 240$

$x^2 + 8x = 240$

Rearrange the equation:

$x^2 + 8x - 240 = 0$

Solve the equation to find the value of $x$.

Solving by splitting the middle term, two numbers whose product is -240 and their sum is 8. The numbers are -12 and 20.

Our equation becomes:

$x^2 + 20x - 12x - 240 = 0$

$x(x + 20) - 12(x + 20) = 0$

Either $(x - 12) = 0$ or $(x + 20) = 0$

$x = 12$ or $x = -20$

Since we don't have negative dimensions, the width is 12 cm and the length is $12 + 8 = 20$ cm.

Therefore, the rectangular plot has the length of 20 cm and the width of 12 cm.

Example 4

A piece of wire 40 cm long is cut into two parts and each part is then bent into a square. If the sum of the areas of these squares is 68 square centimeters, find the lengths of the two pieces of wire.

Solution

Let the length of one piece of wire be $x$ cm.

Then the other piece will be $40 - x$ cm.

When bent into squares:

• Side of first square = $\frac{x}{4}$ cm
• Side of second square = $\frac{40 - x}{4}$ cm

Areas of the squares:

• Area of first square = $\left(\frac{x}{4}\right)^2 = \frac{x^2}{16}$
• Area of second square = $\left(\frac{40 - x}{4}\right)^2 = \frac{(40 - x)^2}{16}$

Given that the sum of the areas is 68 cm²:

$\frac{x^2}{16} + \frac{(40 - x)^2}{16} = 68$

Multiply both sides by 16:

$x^2 + (40 - x)^2 = 1088$

$x^2 + 1600 - 80x + x^2 = 1088$

$2x^2 - 80x + 1600 - 1088 = 0$

$2x^2 - 80x + 512 = 0$

Divide by 2:

$x^2 - 40x + 256 = 0$

Solve by factorization:

$(x - 32)(x - 8) = 0$

$x = 32$ or $x = 8$

Therefore, the two pieces of wire are 32 cm and 8 cm.