Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
Concept, Theory & Solution Roadmap
1. Standard Form of Quadratic Polynomial
A quadratic polynomial is written as: \[ ax^2 + bx + c \quad (a \neq 0) \]
2. Relationship Between Zeroes and Coefficients
If zeroes are \( \alpha \) and \( \beta \), then:
\[ \alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a} \]
3. Solution Roadmap (Exam Strategy)
- Identify \(a, b, c\)
- Compute \(a \times c\)
- Find two numbers → product = \(ac\), sum = \(b\)
- Split middle term
- Factorise using grouping
- Find roots
- Verify using formulas
4. Graph Interpretation (Conceptual Understanding)
Zeroes are the x-intercepts of the graph:
-
\(x^2 - 2x - 8\)
Step 1: Identify coefficients
\[ a=1,\quad b=-2,\quad c=-8 \]Step 2: Compute product
\[ a \times c = 1 \times (-8) = -8 \]Step 3: Find numbers
\[ 2 \text{ and } -4 \quad (product=-8,\ sum=-2) \]Step 4: Split middle term
\[ x^2 - 2x - 8 = x^2 + 2x - 4x - 8 \]Step 5: Factorisation
\[ = x(x+2) -4(x+2) \] \[ = (x+2)(x-4) \]Step 6: Roots
\[ x=-2,\quad x=4 \]Step 7: Verification
\[ \alpha+\beta = 4+(-2)=2 = -\frac{-2}{1} \] \[ \alpha\beta = 4(-2)=-8 = \frac{-8}{1} \] -
\(4s^2 - 4s + 1\)
Step 1
\[ a=4,\ b=-4,\ c=1 \]Step 2
\[ ac = 4 \]Step 3
\[ -2,\ -2 \]Step 4
\[ 4s^2 -2s -2s +1 \]Step 5
\[ 2s(2s-1) -1(2s-1) \] \[ = (2s-1)^2 \]Step 6
\[ 2s-1=0 \Rightarrow s=\frac{1}{2} \]Double root
Verification
\[ \frac{1}{2}+\frac{1}{2}=1,\quad -\frac{-4}{4}=1 \] \[ \frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}=\frac{1}{4} \] -
\(6x^2 - 7x - 3\)
Step 1
\[ a=6,\ b=-7,\ c=-3 \]Step 2
\[ ac=-18 \]Step 3
\[ -9,\ 2 \]Step 4
\[ 6x^2 -9x +2x -3 \]Step 5
\[ 3x(2x-3) +1(2x-3) \] \[ = (3x+1)(2x-3) \]Step 6
\[ x=\frac{3}{2},\quad x=-\frac{1}{3} \]Verification
\[ \frac{3}{2}-\frac{1}{3}=\frac{7}{6}=-\frac{-7}{6} \] \[ \frac{3}{2}\cdot -\frac{1}{3}=-\frac{1}{2}=\frac{-3}{6} \] -
\(4u^2 + 8u\)
Step 1
\[ 4u(u+2) \]Step 2
\[ u=0,\quad u=-2 \]Verification
\[ 0+(-2)=-2=-\frac{8}{4} \] \[ 0 \cdot (-2)=0 \] -
\(t^2 - 15\)
Step 1
\[ t^2=15 \]Step 2
\[ t=\pm \sqrt{15} \]Verification
\[ \sqrt{15}+(-\sqrt{15})=0 \] \[ \sqrt{15}(-\sqrt{15})=-15 \] -
\(3x^2 - x - 4\)
Step 1
\[ ac=-12 \]Step 2
\[ -4,\ 3 \]Step 3
\[ 3x^2 -4x +3x -4 \]Step 4
\[ x(3x-4)+1(3x-4) \] \[ = (x+1)(3x-4) \]Step 5
\[ x=-1,\quad x=\frac{4}{3} \]Verification
\[ \frac{4}{3}-1=\frac{1}{3} \] \[ -\frac{-1}{3}=\frac{1}{3} \] \[ \frac{4}{3}(-1)=-\frac{4}{3} \]
Why This Exercise is Important
For CBSE Board Exams
- Direct 3–4 mark questions from factorisation
- Verification step is often compulsory
- Case-based questions use root relationships
For Competitive Exams (JEE, NDA, SSC, CUET)
- Foundation of quadratic equations
- Used in coordinate geometry & calculus
- Helps in graph-based MCQs
Concept Mastery
- Builds algebraic manipulation speed
- Strengthens number sense
- Improves factorisation accuracy