How to Solve Quadratic Equation? – Tips & Tricks
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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
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