# How to Solve Quadratic Equation? – Tips & Tricks

[email protected] blog

**Dear Aspirants,**

When we deal with Quadratic Equations problem in Quantitative Aptitude, we should use time saving tricks. The ideal & fast way to proceed, including the tricks, is given below:-

**1.** Write down the table (given below) before exam starts, in your rough sheet, for use during exam, Analyse the (+, -) signs in the problem, and refer to the table of signs,

**2.** Write down the new (solution) signs, and see if a solution is obtained instantly. If not, then go to step 3.

**3**. Obtain the two possible values for X & Y, from both the equations,

**4.** Rank the values and get the solution,

__STEP 1__

Firstly, when you enter the exam hall, you need to write down the following **master table** in your rough sheet instantly (only the signs):-

**NOTE:** The above table should be entered before the exam starts, in rough sheet. Because, this will save time, and you can refer to it instantly later on.

__STEP 2__

Try solving the question instantly if possible, by checking whether + or – values can bring us to a conclusion.

**See an example No. 1:- **

x^{2} + 7x + 12 = 0

y^{2} – 5y + 6 = 0

The signs of X’s equation are + and +, which means their **solution is – **and** –**. Both negative values. **(Refer to the table**)

The signs of Y’s equation are – and +, which means their solution values are + and +. **Both positive values.**

Obviously, X’s possible values are both negative… **And Y’s possible values are both positive.**

Obviously, **solution is X < Y.** We found out just by looking at the signs which will take hardly 5 seconds.

**So, any question with signs as “+, +” and “-, +” can be instantly solved**, (unless its an unsolvable exception like x^{2} – 6x + 16 = 0, which can have no real integer as answer. But this is a very rare occasion).

**See another example, Example No. 2:-**

x^{2} + 21x – 32 = 0

y^{2} + 7y – 12 = 0

The question’s signs are + and -, for both X & Y equations. Which means X and Y’s values can be both positive or negative, as per master table…. Answer is an instant CND (Can Not Determine). This takes hardly 5 seconds.

Similarly, any question’s 2 equations with all 4 symbols as negative (-, -), will have each variable’s value either a positive or negative, **so its again a CND.**

__STEP 3__

If you visit Step 3, it means the question is not an instantly solved in Step 2. Here, we write the 2 possible signs of each variable anyways, as plus or minus, in our rough sheet.

We must find the possible values of the variables. Remember, from the 2 symbols (+ or -) derived from the master table, the first symbol connects to the bigger value, and second symbol connects to the smaller value, numerically.

**Example 3:-**

x^{2} – 12x + 32 = 0

y^{2} – 7y + 12 = 0

The question’s symbols are -, + which means both values will be positive for each variable, x and y.

x^{2} – 4x – 8x + 32 = 0

X (x – 4) – 4 (x – 8) = 0

X = 4, X = 8

y^{2} – 7y + 12 = 0

y^{2} – 4y – 3y + 12 = 0

y (y – 4) – 3 (y – 4) = 0

Y= 4, Y = 4

So, X >= Y

**Example 4 is given below:-**

2x^{2} + 3x – 9 = 0

2y^{2} – 11y + 15 = 0

+, – means X will be (-) or (+).

-, + means Y will be (+) or (+)

We have to solve it as we can’t decide instantly.

Now, due to a variation of a factor attached to squared values, the change in method is that **we’ll have to multiply variable less value (9) with factor of the squared variable (2), to get 18 for X’s equation.**

**We factorize 18 as 2 x 3 x 3** and reduce it to 2 values **(which add/subtract to 3) which is 6 x 3**…. Now, divide both values by 2, because that is factor of x^{2}.

So, x = -6/2 or +3/2… In other words, x = -3 or 1.5.

Similarly for Y, we know the signs are + and -, which means both values are positive.

And 15 x 2 = 30. (Again, this is done because y^{2} has a multiplying factor of 2)

**30 = 2 x 3 x 5.** (we’ll have to multiply 2 values of these, to arrive at final 2 values which add upto 11, which are 6 and 5)

11 = 6 + 5. Now, Divide both 6 and 5 by 2 each, to get respective possible values of y.

**So, x = -6/2, 5/2**

**y = 3 or 2.5.**

__STEP 4__

Rank all the 4 values carefully, considering – as lower values and + as higher, as they normally are in mathematics.

**For example,**

**“-5” will get rank 4,**

**“-0.8” will get rank 3,**

**“0.4” will get rank 2,**

**and “4” will get rank 1.**

Write the ranks in your rough sheet, besides the values. Check the ranks and see what conclusion can be derived.

If there’s no tie up/ draw among the ranks, then the answer is easy to find…

So, as you can see from the table above, when all 4 values are different, and if the ranks 1 and 2 are together, they make it a senior variable, otherwise, we can’t determine and it becomes a CND.

When they are tied up, then we’ll have to work it out on rank to rank comparison basis to find a final solution, like below. But this we do only in mind:-

**Let’s carry Example 3 from the Step 3 mentioned before…**

X = -8, -4

Y = -4, -3

X’s ranks would be 4,2

Y’s ranks would be 2,1. (assuming that ranks 2 & 3 are combined to get rank 2)

**We compare them by “Best vs Worst” method this way:-**

**If we ever have to combine a X>Y with X<Y**, then its obviously** CND**. Rest you guys can figure out. Remember to do this Step 4 **in mind itself**, writing only the ranks on rough sheet and nothing else. Don’t make such tables there when the clock is running out…

**Let’s solve Example 4 further from where we left:-**

x = -6/2, 5/2

y = 3 or 2.5.

X’s ranks are= 4, 2

Y’s ranks are= 1, 2.

We compare best to worst, and solve it in mind itself to get Y >= X.

Now let us check some special cases of inequality.

**SPECIAL CASES:-**

**1. SQUARES – **Let us take few examples of non equation values. See example No. 5:-

x^{2} = 1600

y^{2} = 1600

**Answer is CND. Because,** a square would always mean possibility of a positive as well as negative value,

x = 40, -40

y = 40, -40

Hence, any 2 squares or their variations are always CND, irrespective of the values.

** ****NOTE:** Root of a positive value is always positive. If you have to root a negative value, then the question is wrong, and hence, answer is a CND.

**See example No. 6:-**

x^{2} – 243 = 468

y^{2} + 513 = 1023

Don’t touch the pen. Its a CND. Because, ultimately we’ll get a perfect/ imperfect square, which is always a CND.

**2. LINEAR MIXED VARIABLE EQUATIONS**

**Example 7:-**

5x + 4y = 82

4x + 2y = 64

One way to proceed is to equate either the “X”or “Y” part of both equations by multiplication any one equation with a positive integar.

Here, best way would be to double the second equation to form 8x + 4y = 128. We can now substract second equation from first,

8x + 4y = 128

– 5x + 4y = 82

————————

3x = 36

removing 4y from both sides easily to get X=12,

4y = 82 – 5x = 82 – 60 = 22,

y = 5.5,

There’s mostly no positive negative confusion in these cases. So, X > Y is right here.

In PO & SO examinations in the past, we’ve had 5-10 questions on this topic. At the end, remember to practice more and more on this topic, because it can score some badly needed or easily earned scores within seconds.

**Thanks**