Question 10

If one of the zeroes of a quadratic polynomial of the form $x^2+ax+b$ is the negative of the other, then it

Solution:

To solve the problem step by step, we need to analyze the given quadratic polynomial $${\mathrm{x2}}^{}+ax+b$$ and the condition that one of its zeroes (roots) is the negative of the other.

1. Identify the Roots:
Let the roots of the polynomial be $$\alpha$$ and $$\beta$$. According to the problem, one root is the negative of the other. We can express this as:
$$\alpha =-\beta$$

2. Use the Sum of Roots Formula:
For a quadratic polynomial $${\mathrm{x2}}^{}+ax+b$$, the sum of the roots is given by:
$$\alpha +\beta =-\frac{\mathrm{a/}}{1}=-a$$
Substituting $$\alpha =-\beta$$ into the sum of roots, we get:
$$-\beta +\beta =-a⟹0=-a$$
This implies:
$$a=0$$

3. Use the Product of Roots Formula:
The product of the roots for the polynomial is given by:
$$\alpha \cdot \beta =\frac{\mathrm{b/}}{1}=b$$
Substituting $$\alpha =-\beta$$ into the product of roots, we have:
$$(-\beta )\cdot \beta =b⟹-{\mathrm{\beta 2}}^{}=b$$
Rearranging gives us:
$${\mathrm{\beta 2}}^{}=-b$$

4. Analyze the Implications:
Since $${\mathrm{\beta 2}}^{}$$ is a square of a real number, it must be non-negative:
$${\mathrm{\beta 2}}^{}\ge 0$$
Therefore, for $$-b$$ to be non-negative, we must have:
$$-b\ge 0⟹b\le 0$$

5. Conclusion:
From the above analysis, we conclude that:
– The coefficient $$a$$ must be $$0$$ (indicating there is no linear term).
– The constant term $$b$$ must be less than or equal to $$0$$ (indicating that it is either negative or zero).

Thus, the final conclusion is that the polynomial has no linear term and the constant term is negative or zero.