# fast-track mathematics
1. The graph of a quadratic polynomial $p(x)$ passes through the points $(-6,0),(0,-30)$, $(4,-20)$ and $(6,0)$. The zeroes of the polynomial are:
(a) $-6,0$
(b) 4,6
(c) $-30,-20$
(d) $-6,6$
[CBSE SQP 2024]
Ans. (d) $-6,6$
Explanation: The quadratic polynomial passes through the points $(-6,0)$ and $(6,0)$, so its zeroes are $x=-6$ and $x=6$.
2. A quadratic polynomial having zeroes $-\sqrt{\frac{5}{2}}$ and $\sqrt{\frac{5}{2}}$ is:
(a) $x^2-5 \sqrt{2} x+1$
(b) $8 x^2-20$
(c) $15 x^2-6$
(d) $x^2-2 \sqrt{5} x-1$
[CBSE SQP 2024]
Ans. (b) $8 x^2-20$
Explanation: If $x=-\sqrt{\frac{5}{2}}$ and $x=\sqrt{\frac{5}{2}}$ are zeroes of the quadratic polynomial, then $\left(x+\sqrt{\frac{5}{2}}\right)$ and $\left(x-\sqrt{\frac{5}{2}}\right)$ are factors so their product will give us the quadratic polynomial. $\therefore k\left(x+\sqrt{\frac{5}{2}}\right)\left(x-\sqrt{\frac{5}{2}}\right)$, where $k$ is any arbitrary constant.
$$
\begin{aligned}
\Rightarrow k\left[(x)^2-\left(\sqrt{\frac{5}{2}}\right)^2\right] & {\left[\because(a+b)(a-b)=a^2-b^2\right] } \\
& =k\left[x^2-\frac{5}{2}\right] \\
& =k\left[\frac{2 x^2-5}{2}\right] \\
& =\frac{k}{2}\left(2 x^2-5\right)
\end{aligned}
$$
For $k=8$ quadratic polynomial is:
$$
\begin{aligned}
& =\frac{8}{2}\left(2 x^2-5\right) \\
& =4\left(2 x^2-5\right) \\
& =8 x^2-20
\end{aligned}
$$
