How To Solve QUADRATIC EQUATION Short Tricks
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The quadratic equation is very important for all competitive exams generally 4-5 questions come from this topic in maximum exams. So we are here providing you the concepts and important short tricks to solve quadratic equation in very fast and efficient way. At the end, we will provide few practice questions also apply the trick on those and see that you have got the trick or not.
So the first thing that comes to our mind is what is a quadratic equation.
The Equation that is in the form of a(x)^2 + bx + c =0 is known as a quadratic equation.
where x represents an unknown, and a, b, and c represent known numbers such that a is not equal to 0. If a = 0, then the equation is linear, not quadratic. The numbers a, b, and c are the coefficients of the equation and may be distinguished by calling them, respectively, the quadratic coefficient, the linear coefficient, and the constant or free term.
In the equation, we have two equation in quadratic form and we have to find their roots and compare them.
Let the roots be as x1, x2, y1, y2 and then we can compare them by following method.
It means we compare ‘x’ with both factors of ‘y’ i.e. y1, y2, then x2 with both the factors of ‘y’ and answer according to it. We use the sign in the equation to find the sign of roots and the forms of an equation to find the magnitude of roots.
| Sign of coefficient of ‘x’ | Sign of coefficient of ‘y’ | Signs of roots | |
| + | + | – | – |
| + | – | – | + |
| – | + | + | + |
| – | – | + | – |
Example
X^2 – 7x + 12 = 0 y^2 + y – 20=0
By using table, Roots are
X1 x2 y1 y2
+4 +3 -5 +4
In this case we can see that X2 > Y1
X1 = X2 & Y2 > X2
So we cannot determine the relationship so answer will be CND (cannot determine). Below is the chart where you can directly answer after getting the roots.
| X1 | X2 | Y1 | Y2 | Results | |
| Case 1 | +5 | +4 | +2 | -1 | X>Y |
| Case 2 | +5 | +4 | +4 | +1 | X>=Y |
| Case 3 | +3 | +4 | +4 | +1 | Y>X |
| Case 4 | +3 | +5 | +5 | +7 | Y>=X |
| Case 5 | +9 | +6 | +7 | +4 | CND |
| Case 6 | +4 | +7 | +9 | +6 | CND |
| Case 7 | +8 | +5 | +4 | +8 | CND |
| Case 8 | +8 | +4 | +5 | +8 | CND |
In some special cases by using sign of quadratic equation, we can answer directly just comparing signs of roots by using table
Type 1 : If one equation have sign “-“,”+” and other has “+”,”+”
Type 2 : If both equations have sign “+”,”-“
Type 3 : If both equations have sign “-“,”-“
Type 4 : If one equation have sign “+”,”-“ and other has sign “-“,”-“
Type 1 : When one equation has positive roots and other equation has negative roots the answer will be roots of the positive equation is greater than negative ones.
Example : x^2 – 7x + 10 = 0
Y^2 + 8y + 15 = 0
By just comparing sign of equation by sign root using table
Here Roots are +x1, +x2, -y1, -y2
So, we can see that both roots of “x” are positive & roots of “y” is negative.
Therefore X>Y
Type 2 : In this type of equation, roots of the equation is positive & negative. Here by comparing the roots i.e. x2 is greater than y1 & x1 is less than y2 so we cannot determine the answer, the answer will be CND.
Example : x^2 + x – 56 = 0
y^2 + 2y – 15 = 0
By just comparing sign of equation by sign root using table
Roots are -x1, +x2, -y1, +y2
Here +X2 > -Y1 & -X1 < +Y2
Therefore answer is CND (Cannot determine)
Type 3 : In this type of equation, roots of the equation is positive & negative. Hereby comparing the roots i.e. x1 is greater than y2 & x2 is less than y1 so we cannot determine the answer, the answer will be CND.
Example : x^2 – x – 6 = 0
y^2 + 2y – 15 = 0
By just comparing sign of equation by sign root using table
Roots are +x1, -x2, +y1, -y2
Here +X1 > -Y2 & -X2 < +Y1
Therefore answer is CND (Cannot determine)
Type 4 : In these type of equations, roots of the equation is positive & negative. Hereby comparing the roots i.e. x2 is greater than y2 & x1 is less than y1 so we cannot determine the answer, the answer will be CND.
Example: x^2 + x – 56 = 0
20(y)^2 – y – 12 = 0
By just comparing sign of equation by sign root using table
Roots are -x1, +x2, +y1, -y2
Here X1 < Y1 & X2 > Y2
Therefore relation cannot be established so the answer is CND.
Directions: In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly:
(i) p2 – 5p +6 = 0
(ii) q2-2q+1 = 0
- If p> q
- If p < q
- If p ≥ q
- If p ≤ q
- If p = q or relation cannot be established
Ans : = A
Roots are p = +3, +2
q = +1,+1
Here p > q
(ii) 15p2 + 5p + 1 = 0
(ii) 2q2 + 10q – 48 = 0
- If p > q
- If p < q
- If p ≥ q
- If p ≤ q
- If p = q or relation cannot be established
Ans : = E
Roots are p = -0.2, -0.33
q = -8, +3
Here no relations are formed between p & q
iii) 6p2 + p – 1 = 0
8q2 + 10q + 3 = 0
- If p > q
- If p < q
- If p ≥ q
- If p ≤ q
- If p = q or relation cannot be established
Ans : = C
Roots are p = -0.5, 0.33
q= -0.75, -0.5
Here roots of p are greater than equal to q.
iv) 4p2– 9p – 9 = 0
(ii) 3q2 + 2q – 21 = 0
- If p > q
- If p < q
- If p ≥ q
- If p ≤ q
- If p = q or relation cannot be established
Ans : E
Roots are p = -0.75, 3
q = -3, 2.3
Here no relations are formed between p & q
