Quadratic Equations – Study Notes (TS EAMCET – Mathematics)
Topic: Quadratic Equations
Exam: TS EAMCET
Level: 10‑12 (High School)
Duration: 25 min (for the mathematics section)
1. Simple Explanation
A quadratic equation is a polynomial equation of degree 2:
[ ax^{2} + bx + c = 0 \qquad (a\neq 0) ]
The unknown (x) is called a root or solution of the equation.
In the TS EAMCET you will encounter:
- Factoring (when the quadratic is factorable over integers/rationals)
- Quadratic formula (when factoring is not obvious)
- Completing the square (to get vertex form or to solve quickly)
- Discriminant ( \Delta = b^{2} - 4ac ) (to decide the nature of roots)
- Vieta’s relations (sum and product of roots)
2. Key Points
| Item | What to remember | Quick Check |
|---|---|---|
| Standard form | (ax^{2}+bx+c=0) | Coefficients (a,b,c) are integers/rationals |
| Discriminant | (\Delta=b^{2}-4ac) | (\Delta>0): two distinct real roots (\Delta=0): one real (repeated) root (\Delta<0): two complex roots |
| Quadratic formula | (x=\dfrac{-b\pm\sqrt{\Delta}}{2a}) | Use when (\Delta) is a perfect square for integer roots |
| Factorisation | ((px+q)(rx+s)=0) → (p r = a,; q s = c,; ps+qr=b) | Check if (c) is small or a perfect square |
| Completing the square | (ax^{2}+bx+c = a\bigl[(x+\tfrac{b}{2a})^{2} - (\tfrac{b^{2}}{4a^{2}}- \tfrac{c}{a})\bigr]) | Gives vertex (\bigl(-\tfrac{b}{2a},; \tfrac{4ac-b^{2}}{4a}\bigr)) |
| Vieta’s formulas | For roots (r_1,r_2): (r_1+r_2 = -\dfrac{b}{a}) (r_1r_2=\dfrac{c}{a}) |
Handy for checking answers |
| Graph of (y=ax^{2}+bx+c) | Parabola opening up ((a>0)) or down ((a<0)) | Vertex gives max/min, axis of symmetry (x=-\tfrac{b}{2a}) |
| TS EAMCET tips | 1. Check for easy factorisation before using the formula. 2. If (\Delta) is not a perfect square, write answer in radical form. 3. Use Vieta’s to confirm roots quickly. 4. Pay attention to signs (especially when (a<0)). |
3. Worked Example
Solve
[
3x^{2}-12x+9 = 0
]
Step‑by‑Step
| Step | Action | Result |
|---|---|---|
| 1 | Identify (a,b,c). | (a=3,; b=-12,; c=9) |
| 2 | Check factorability: look for two numbers whose product (ac=27) and sum (-12). | (-9) and (-3) give ((-9)(-3)=27) and (-9-3=-12) → factorable. |
| 3 | Factor: | |
| [ | ||
| 3x^{2}-12x+9 = 3(x^{2}-4x+3)=3(x-3)(x-1) | ||
| ] | ||
| 4 | Set each factor to zero. | (x-3=0 \Rightarrow x=3) (x-1=0 \Rightarrow x=1) |
| 5 | Write solution set. | (\boxed{x=1,;3}) |
| 6 | Optional – verify with quadratic formula: | |
| (\Delta = (-12)^{2}-4(3)(9)=144-108=36=6^{2}) | ||
| (x=\dfrac{-(-12)\pm 6}{2\cdot 3}=\dfrac{12\pm 6}{6}) → (x=3,;1). | Same result |
Quick Check (Vieta)
Sum of roots (;1+3=4 = -b/a = 12/3=4) ✔
Product of roots (;1\cdot3=3 = c/a = 9/3=3) ✔
4. Quick Revision Summary
| Concept | Remember |
|---|---|
| Standard form | (ax^{2}+bx+c=0,;a\neq0) |
| Discriminant | (\Delta=b^{2}-4ac) – tells the type of roots |
| Quadratic formula | (x=\dfrac{-b\pm\sqrt{\Delta}}{2a}) |
| Factorisation | Find two numbers ((p,q)) s.t. (pq=ac,;p+q=b) |
| Vieta’s | (r_{1}+r_{2}=- |