Tips & Tricks to Solve Quadratic Equations
Shortcut tricks are very important things in competitive exams.Time is the main factor in the competitive exams.If you know how to manage time then you will surely do great in your exam.
Quantitative section in competitive exam is the most important part of the exam.It doesn’t mean that other topics are not so important.You can get a good score only if you get a good score in Quantitative section.You can only get good marks by practicing more and more.
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:-
- 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,
- Write down the new (solution) signs, and see if a solution is obtained instantly. If not, then go to step 3.
- Obtain the two possible values for X & Y, from both the equations,
- Rank the values and get the solution.
Firstly, when you enter the exam hall, you need to write down the following master table in your rough sheet instantly (only the signs):-
Try solving the question instantly if possible, by checking whether + or – values can bring us to a conclusion.
x2 + 7x + 12 = 0
y2 – 5y + 6 = 0
The signs of X’s equation are + and +, which means their solution is – and –. Both negative values.
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 x2 – 6x + 16 = 0, which can have no real integer as answer. But this is a very rare occasion).
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.
x2 – 12x + 32 = 0
y2 – 7y + 12 = 0
The question’s symbols are -, + which means both values will be positive for each variable, x and y.
x2 – 4x – 8x + 32 = 0
X (x – 4) – 4 (x – 8) = 0
X = 4, X = 8
y2 – 7y + 12 = 0
y2 – 4y – 3y + 12 = 0
y (y – 4) – 3 (y – 4) = 0
Y= 4, Y = 4
So, X >= Y
Rank all the 4 values carefully, considering – as lower values and + as higher, as they normally are in mathematics.
“-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.
FEW SPECIAL CASES:-
Let us take few examples of non equation values.
x2 = 1600
y2 = 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.
LINEAR MIXED VARIABLE EQUATIONS
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,
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.