Daily Practice Problems

📝 SolutionFor a quadratic $ax^2 + bx + c$, sum of zeroes $\alpha + \beta = -b/a$ and product $\alpha\beta = c/a$. Here $a=1, b=-5, c=6$, so $\alpha + \beta = 5$ and $\alpha\beta = 6$. Therefore $\alpha + \beta + \alpha\beta = 5 + 6 = 11$. The key skill is reading off the coefficient relationships directly instead of solving for the roots — it saves time in the board exam.

📝 SolutionThe number of times the graph of a polynomial cuts the x-axis equals the number of real zeroes, and a polynomial of degree $n$ has at most $n$ zeroes. If the graph crosses the x-axis at 3 distinct points, the polynomial has at least 3 real zeroes, so its degree must be at least 3. The minimum degree is therefore 3. A common mistake is confusing 'number of turning points' with 'degree' — always count the x-axis intersections for the number of zeroes.

📝 SolutionIf the zeroes are reciprocals, then $\alpha$ and $1/\alpha$, so their product $\alpha \cdot \tfrac{1}{\alpha} = 1$. Product of zeroes $= c/a = k/2$. Setting $k/2 = 1$ gives $k = 2$. Whenever a question says 'one zero is the reciprocal of the other', immediately use product of zeroes $= 1$ — that is the fastest route.

📝 SolutionA quadratic with given sum $S$ and product $P$ of zeroes is $x^2 - Sx + P$. Here $S = -3$ and $P = 2$, so the polynomial is $x^2 - (-3)x + 2 = x^2 + 3x + 2$. Watch the sign carefully: the coefficient of $x$ is $-S$, so a negative sum of zeroes produces a positive middle coefficient.

📝 SolutionAssertion (A) is a standard theorem: a polynomial of degree $n$ has at most $n$ zeroes. Reason (R) is also true — each real zero corresponds to a point where the graph meets the x-axis, so a degree-$n$ graph meets the x-axis at most $n$ times. Since the geometric statement (R) is precisely why the algebraic bound (A) holds, R is the correct explanation of A. Both true, R explains A.