Linear equations have always been the easiest because even though there is a set of equations yet each variable will have only one solution but GRE tests your ability to solve quadratic equations as well. Quadratic equations can appear in in front of in the form of parabolic equations or other coordinate geometry equations. Knowing how to solve the equations of quadratic can come handy and reduce many unwanted complexities, means saving a lot of your time.

So, it is pretty clear now that quadratic equation is an important topic in GRE and is linked with many other topics. Hence, before you jump on solving complex quadratic equations, let’s first understand what 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: \(x^{2} + 8x = -12, \; p^{2} = p, \; \frac{4}{x} = x + 3…\) 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 \((-12) \times (-12) = 144\)

Now Solve, \(12^{2}\) which will be equal to \((12) \times (12) = 144\)

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 produce 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 ‘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 ‘a.’

This is where quadratic comes in view.

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

How to solve these types of quadratic equations?

Solve \(x^{2} + 5x – 6 = -12\), and find the possible values of x.

The first step involves of setting the equation to zero.

So, it becomes, \(x^{2} + 5x + 6 = 0\)

Rewrite the quadratic equation in factor form: \(x^{2} + 5x + 6 = (x + … )(x + … )\)

Determine the values to fill the slots, so that on adding these value it yields ‘b’ or in our example 5 and on multiplying these values it yields ‘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:

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

or

\(x = -2 \; or \; x = -3\)Alternatively, in place of thinking the values, you can use the standard formula of roots, which is

\(x = \frac{-b \pm \sqrt{b^{2} – 4ac}}{2a}\)Quadratic formulas to memorize

\(x^{2} – y^{2} = (x + y)(x – y)\)

\((x – y)^{2} = x^{2} – 2xy + y^{2}\)

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

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

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