QUADRATIC EQUATION – 3 METHODS, WORD PROBLEMS, DERIVATION

A quadratic equation is one in which the highest power of the unknown is two.

To solve a quadratic equation means to determine the root of the equation.

The roots of a quadratic equation are the numerical value that makes the equation true.

There are three methods through which we can solve quadratic equations, namely factorization method, completing the square method and formula method.

Factorization Method

This is arguably the easiest and the commonest method of solving quadratic equations.

As its name implies, the factorization method is the method of factoring the left-hand side of a quadratic equation and then equating the result to zero.

Example 1

Solve $2x^2-3x-2=0$

Solution:

By factorization,

$2x^2-3x-2=0$

$2x^2-4x+1x-2=0$

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

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

Now, we equate each result to zero

$2x+1=0$ or $x-2=0$

$2x=-1$ or $x=2$

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

Example 2

Determine the roots of $6x^2-7x-5=0$

Solution:

$6x^2-10x+3x-5=0$

$2x(3x-5)+1(3x-5)=0$

$(2x+1)(3x-5)=0$

Again, we equate the results to 0

$2x+1=0$ or $3x-5=0$

$2x=-1$ or $3x=5$

$x=\frac{-1}{2}$ or $x=\frac{5}{3}$

Completing The Square Method

To solve the quadratic equation, here are the procedure.

1. Rearrange the equation so that all the x's on are one side while the constant is on the other side

2. Add the square half of the coefficient of x to both sides

3. Express the left-hand side as a perfect square

4. Add square root to both sides

Example 3

Solve $x^2+25x-150=0$ using completing the square method

Solution:

First, we move the constant to the other side of the equal sign

$x^2+25x=150$

Add the square half of the coefficient of x to both sides

$x^2+25x+(\frac{25}{2})^2=150+(\frac{25}{2})^2$

$x^2+25x+(\frac{25}{2})^2=150+\frac{625}{4}$

$x^2+25x+(\frac{25}{2})^2=\frac{600+625}{4}$

$x^2+25x+(\frac{25}{2})^2=\frac{1225}{4}$

The left-hand side is a perfect square. Hence

$(x+\frac{25}{2})^2=\frac{1225}{4}$

Next, we add square root to both sides

$\sqrt{(x+\frac{25}{2})^2}=±\sqrt{\frac{1225}{4}}$

$x+\frac{25}{2}=±\frac{35}{2}$

$x=±\frac{35}{2}-\frac{25}{2}$

Splitting the plus and minus

$x=+\frac{35}{2}-\frac{25}{2}$ or $-\frac{35}{2}-\frac{25}{2}$

$x=\frac{35}{2}-\frac{25}{2}$ or $-\frac{35}{2}-\frac{25}{2}$

$x=\frac{10}{2}-$ or $-\frac{60}{2}$

$x=5$ or $-30$

Formula Method

We can also solve quadratic equations using the formula method.

The formula is

$$x=\frac{-b±\sqrt{b^2-4ac}}{2a}$$

Example 4

Solve $x^2-4x-192=0$

Solution:

Let $a=1$, $b=-4$, $c=192$

$x=\frac{-(-4)±\sqrt{(-4)^2-4(1)(-192)}}{2(1)}$

$x=\frac{4±\sqrt{16+768}}{2}$

$x=\frac{4±\sqrt{784}}{2}$

$x=\frac{4±28}{2}$

Splitting the plus and minus sign

$x=\frac{4+28}{2}$ or $\frac{4-28}{2}$

$x=\frac{32}{2}$ or $\frac{-24}{2}$

$x=16$ or $-12$

Example 5

Solve $30x^2-130x-100$

Solution:

Here, a=30, b=-130, c=-100

$x=\frac{-(-130)±\sqrt{(-130)^2-4(30)(-100)}}{2(30)}$

$x=\frac{130±\sqrt{16900+12000}}{60}$

$x=\frac{130±\sqrt{28900}}{60}$

$x=\frac{130±170}{60}$

$x=\frac{130+170}{60}$ or $\frac{130-170}{60}$

$x=\frac{300}{60}$ or $\frac{-40}{60}$

$x=5$ or $\frac{-2}{3}$

I recommend that you read this post to solidify your understanding of quadratic equations using the formula method.

Sometimes, quadratic equations are expressed in words. These are called word problems

Example 6

Three years ago, a father was four times as old as his daughter is now. The product of their present age is 430. Calculate the present age of the daughter and father

Solution:

Let the father's age be y.

Let the daughter's age be x.

From the first sentence,

$y-3=4(x)$

$y-3=4x$

$y=4x+3$

From the second sentence,

$xy=430$

Recall that $y=4x+3$

$x(4x+3)=430$

$4x^2+3x=430$

$4x^2+3x-430$

By factorization

$4x^2+43x-40x-430$

$x(4x+43)-10(4x+43)=0$

$(x-10)(4x+43)=0$

$x=10$ or $4x=-43$

$x=10$ or $x=\frac{-43}{4}$

More word problems are available here

So far, we have solved quadratic equations where the real equations are given.

However, sometimes, you may be asked to obtain a quadratic equation from the root of the equation.

Related posts

Example 7

Find the quadratic equation whose root is $\frac{1}{2}$ and $-7$.

Solution:

$x=\frac{1}{2}$ and $x=-7$

$x-\frac{1}{2}=0$ and $x+7=0$

$(x-\frac{1}{2})(x+7)=0$

Expanding the brackets

$x(x+7)-\frac{1}{2}(x+7)$

$x^2+7x-\frac{x}{2}-\frac{7}{2}=0$

To clear the fraction, we multiply through by 2

$2(x^2)+2(7x)-2(\frac{x}{2})-2(\frac{7}{2})=2(0)$

$2x^2+14x-x-7=0$

$2x^2+13x-7=0$

We have discussed the derivation of a quadratic equation from a quadratic root here

Voila, you have just learned the quadratic equation. Are you enjoying what you're reading?

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