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Polynomials
NCERT Exemplar Ch 02
CBSE_NCERT_EXEMPLAR_CH02
Grade 10
Question:
If the zeroes of the quadratic polynomial $x^2 + (a + 1)x + b$ are $2$ and $-3$, then:
$a = -7, b = -1$
$a = 5, b = -1$
$a = 2, b = -6$
$a = 0, b = -6$
$a = -7, b = -1$
$a = 5, b = -1$
$a = 2, b = -6$
$a = 0, b = -6$
Step-by-Step Solution
Key Concept: Sum of zeroes $= -\dfrac{\text{coeff of } x}{\text{coeff of } x^2}$ and Product of zeroes $= \dfrac{\text{constant term}}{\text{coeff of } x^2}$.
Stepwise Solution:
Sum of zeroes: $2 + (-3) = -(a + 1) \Rightarrow -1 = -a - 1 \Rightarrow a = 0$. [0.5 Mark]
Product of zeroes: $(2)(-3) = b \Rightarrow b = -6$. [0.5 Mark]
Marking Scheme:
• Using sum of zeroes to find $a$: 0.5 Mark
• Using product of zeroes to find $b$: 0.5 Mark
Sum of zeroes: $2 + (-3) = -(a + 1) \Rightarrow -1 = -a - 1 \Rightarrow a = 0$. [0.5 Mark]
Product of zeroes: $(2)(-3) = b \Rightarrow b = -6$. [0.5 Mark]
Marking Scheme:
• Using sum of zeroes to find $a$: 0.5 Mark
• Using product of zeroes to find $b$: 0.5 Mark
Correct Answer: $a = 0, b = -6$
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