Linear equations have always been the easiest because even though there are a set of equations yet each variable will have only one solution. However the GRE exam tests your ability to solve quadratic equations as well. Quadratic equations can appear in the form of parabolic equations or other coordinate geometry equations. If one knows how to solve quadratic equations then this can come in handy and reduce many unwanted complexities, since it saves a lot of your time.

So, it is pretty clear now that quadratic equations is an important topic in the GRE test and is linked with many other topics. Hence, before you jump to solve complex quadratic equations, let’s first understand what a quadratic equation is:

An equation in which at least one of the variables is raised to an even power is said to be a quadratic equation.

**For example:** x^{2} + 8x = -12, p^{2}= p, 4x=x+3â€¦. etc. all these are various forms of quadratic equations.

To understand these concepts clearly, let us take an example.

Evaluate; (-12)^{2}

Using the BODMAS we know that the answer will be (-12) Ã— (-12) =144

Now solve 12^{2}, which will be equal to (12) Ã— (12) =144

It can be easily noted that 12 and -12 both produce the same result, and the reason is, if a variable is raised to an even power, it always produces a positive value irrespective of the number being positive or negative.

Now you must be wondering where quadratic comes in all these because these are the basic concepts of exponents. Right?

If you want to get the answer to this question then first solve this problem:

If a^{2} = 144, then find the value of â€˜a.’

You might be tempted just to square root 144 and give the answer that a = 12, but you are missing out the above concept, the concept of even power. Hence, 12 and -12 both are the value of â€˜a.’

This is where quadratic comes in view.

A common form of quadratic equations:

The most common form of quadratic equation is: **ax ^{2}+ bx +c = 0**

How to solve these types of quadratic equations?

Let us understand it through an example

**Example : **Take this equation, x^{2} + 5x+ 6 = 12, and find the possible values of x.

**Factorization Method**

- The first step involves setting the equation to zero.So, it becomes, x
^{2}+ 5x+ 6 = 0 - Rewrite the quadratic equation in factored form: x
^{2}+ 5x + 6 (x )(x ) = 0 - Look at the sign in front of the third term(6). It is a + sign, but multiplications of two negative numbers can also give a positive result. So look for the sign in front of the second term(5x). That is also positive. So place the + sign after x.(x + )(x + ) = 0
- Determine the values to fill the slots, so that on adding these value it yields â€˜b’ or in our example 5 and on multiplying these values it yields â€˜c’ i.e. here it yields 6. So, what those values can be? Yes, it is 2 and 3.

Hence after filling the slots it becomes (x + 2)(x + 3) = 0 - Solve for x. As the rule says; if a product of two or more factors is equal to zero, then at least one of the factor is equal to zero. Hence, either:

(x+ 2)= 0 or (x+3) =0

so, x = -2 or x = -3

Alternatively, in place of thinking the values, you can use the **standard formula of roots**, which is

x = -b Â± b^{2}â€“ 4ac2a

**Quadratic formulas** to memorize:

- \(x^{2} – y^{2} = (x + y)(x – y)\)

\((x –Â y)^{2} = x^{2} – 2xy + y^{2}\)

\((x + y)^{2} = x^{2} + 2xy + y^{2}\)

Solving quadratic will be fun if you start treating it like a game in place of thinking it like a hectic math problem.

Let us solve the same question using the standard formula of roots x^{2} + 5x + 6 = 0

=> -5 Â± âˆš[5^{2} – 4(1*6)]/ 2*1

=> -5 Â± âˆš(25-24)/2

=> (-5 Â± 1)/2

=> -6/2 or -4/2

=> -3 or -2

This is the same answer we got using the factorization method. When the equation becomes complex, it is better to use the formula of roots as given above.

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