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.

**Word Problem On Quadratic Equation**

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

**Derivation Of Quadratic Equation From Quadratic Roots**

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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