Quadratic Equations and Expressions
3.0 Nature of roots
3.0 Nature of roots
Consider the quadratic equation $a{x^2} + bx + c = 0$ having $\alpha {\text{ and }}\beta $ as its roots.
Question 2. Find all values of $a$ for which the quadratic equation $$(a + 1){x^2} + 2(a + 1)x + a - 2 = 0$$
(a) has two distinct roots.
(b) has no roots.
(c) has two equal roots.
Solution: As the equation in quadratic, $$\begin{equation} \begin{aligned} a + 1 \ne 0 \\ a \ne - 1 \\\end{aligned} \end{equation} $$
Discriminant $$\begin{equation} \begin{aligned} D = 4{(a + 1)^2} - 4(a + 1)(a - 2) \\ {\text{ = }}12(a + 1) \\\end{aligned} \end{equation} $$
(a) For two distinct roots, $D>0$ i.e., $a>-1$
(b) For no roots, $D<0$ i.e., $a<-1$
(c) For two equal roots, $D=0$ i.e., $a=-1$ which is not possible and contradicts the hypothesis. So, the equation can not have two equal roots.
Question 3. Find all the integral values of $a$ for which the quadratic equation $$(x - a)(x - 10) + 1 = 0$$ has integral roots.
Solution: The quadratic equation is $${x^2} - (a + 10)x + 10a + 1 = 0$$
Since integral roots will always be rational, which means $D$ should be a perfect square.
$$\begin{equation} \begin{aligned} D = \frac{{{{\left[ { - (a + 10)} \right]}^2} - 4 \times 1 \times (10a + 1)}}{{2 \times 1}} = {(a - 10)^2} - 4 \\ 4 = {(a - 10)^2} - D \\\end{aligned} \end{equation} $$
If $D$ is a perfect square it means we need the difference of two perfect square equal to $4$, which is possible only when $$D=0$$ and$$\begin{equation} \begin{aligned} {(a - 10)^2} = 4 \\ a - 10 = \pm 2 \\ a = 12{\text{ or }}8 \\\end{aligned} \end{equation} $$
