Find the zeroes of the quadratic polynomial x2 + 7x + 10, and verify the relationship between the zeroes and the coefficients.
Step-by-Step Solution
Key Concept: For a quadratic polynomial $ax^2+bx+c=0$, if its zeroes are $\alpha$ and $\beta$, then the sum and product of the zeroes are given by the relationships $\alpha+\beta=-\dfrac{b}{a}$ and $\alpha\beta=\dfrac{c}{a}$. These are derived from factorising the polynomial as $a(x-\alpha)(x-\beta)$.
1. Write the given polynomial: \[f(x)=x^{2}+7x+10\] Here, $a=1$, $b=7$, $c=10$.
2. Find the zeroes by factorisation:
\[x^{2}+7x+10 = (x+5)(x+2)\]
Set each factor to zero:
\[x+5=0 \;\Rightarrow\; x=-5\]
\[x+2=0 \;\Rightarrow\; x=-2\]
Hence the zeroes are $\alpha=-5$ and $\beta=-2$.
3. Verify the sum of zeroes:
\[\alpha+\beta = (-5)+(-2) = -7\]
According to the theorem, \[\alpha+\beta = -\frac{b}{a}= -\frac{7}{1}= -7\]
The calculated sum matches the theoretical value.
4. Verify the product of zeroes:
\[\alpha\beta = (-5)(-2) = 10\]
According to the theorem, \[\alpha\beta = \frac{c}{a}= \frac{10}{1}=10\]
The calculated product also matches the theoretical value.
5. Conclusion: The zeroes $-5$ and $-2$ satisfy both the sum and product relationships with the coefficients of the polynomial.
Correct Answer: The zeroes are $\alpha=-5$ and $\beta=-2$. Their sum $\alpha+\beta=-7 = -\dfrac{b}{a}$ and product $\alpha\beta=10 = \dfrac{c}{a}$, confirming the required relationships.