### Quadratic Equations in One Unknown-Examples

• Given the quadratic equation $$p{x^2} + (3p + 2q)x + 6q = 0$$, where p and q are constants.
(a) Find the discriminant of the equation.
(b) Prove that the equation has real roots.
(c) Hence prove that $$19{x^2} + 81x + 72 = 0$$ has real roots.
• (a)
$$\begin{array}{1}\Delta = {(3p + 2q)^2} - 4(p)(6q)\\ = 9{p^2} + 12pq + 4{q^2} - 24p{q^{^{}}}\\ = 9{p^2} - 12pq + 4{q^2}^{^{}}\\ = {\underline{\underline {{{(3p - 2q)}^2}}} ^{}}\end{array}$$

(b)
$$\Delta = {(3p - 2q)^2} \ge 0$$
∴ The equation has real roots.

(c)
Let $$p = 19$$ and $$q = 12$$, then
$$\begin{array}{1}p{x^2} + (3p + 2q)x + 6q = 0\\19{x^2} + (57 + 24)x + 72 = 0\\19{x^2} + 81x + 72 = 0\end{array}$$
∴ The equation $$19{x^2} + 81x + 72 = 0$$ has real roots.
• If a and b are the roots of the equation $$x(2x + 3) = 1$$ and $$\alpha > \beta$$, find the value of each of the following.
(a) $${(\alpha - \beta )^2}$$
(b) $$\frac{\alpha }{\beta } - \frac{\beta }{\alpha }$$
(c) $$(\alpha + 3)(\beta - 3)$$
• $$\begin{array}{c}x(2x + 3) = 1\\2{x^2} + 3x - 1 = 0\end{array}$$
Sum of roots $$= \alpha + \beta = - \frac{3}{2}$$
Product of roots $$= \alpha \beta = \frac{{ - {\rm{ }}1}}{2} = - \frac{1}{2}$$

(a)
$$\begin{array}{1}{(\alpha - \beta )^2} = {\alpha ^2} - 2\alpha \beta + {\beta ^2}\\ = ({\alpha ^2} + 2\alpha \beta + {\beta ^2}) - 4\alpha \beta \\ = {(\alpha + \beta )^2} - 4\alpha \beta \\ = {( - \frac{3}{2})^2} - 4( - \frac{1}{2})\\ = \frac{9}{4} + 2\\ = \underline{\underline {\frac{{17}}{4}}} \end{array}$$

(b)
$$\begin{array}{1}\frac{\alpha }{\beta } - \frac{\beta }{\alpha } = \frac{{{\alpha ^2} - {\beta ^2}}}{{\alpha \beta }}\\ = \frac{{(\alpha + \beta )(\alpha - \beta )}}{{\alpha \beta }}\end{array}$$
∵ $$\alpha > \beta$$
∴ $$\begin{array}{1}\alpha - \beta > 0\\\alpha - \beta = {\sqrt {{{(\alpha - \beta )}^2}} ^{}}\end{array}$$
∴ $$\begin{array}{1}\frac{\alpha }{\beta } - \frac{\beta }{\alpha } = \frac{{( - {\rm{ }}{\textstyle{3 \over 2}})\sqrt {{\textstyle{{17} \over 4}}} }}{{ - {\rm{ }}{\textstyle{1 \over 2}}}}\\ = \underline{\underline {\frac{{3\sqrt {17} }}{2}}} \end{array}$$

(c)
$$\begin{array}{1}(\alpha + 3)(\beta - 3) = \alpha \beta - 3(\alpha - \beta ) - 9\\ = - \frac{1}{2} - 3\sqrt {\frac{{17}}{4}} - {9^{^{^{^{^{}}}}}}\\ = {\underline{\underline {\frac{{ - {\rm{ }}19 - 3\sqrt {17} }}{2}}} ^{}}\end{array}$$
• It is given that α and β are the roots of the equation $${x^2} + px + q = 0$$, where p and q are constants.
(a) Express the values of the following in terms of p and q.
(i) $$\frac{1}{\alpha } + \frac{1}{\beta }$$ (ii) $$\frac{\alpha }{\beta } + \frac{\beta }{\alpha }$$
(b) Find a quadratic equation in x with roots $$(\frac{2}{\alpha } + \alpha )$$ and $$(\beta + \frac{2}{\beta })$$.
• (a) ∵ α and β are the roots of the equation $${x^2} + px + q = 0$$.
∴ $$\alpha + \beta = - {\rm{ }}p$$ and $$\alpha \beta = q$$
(i)
$$\begin{array}{1}\frac{1}{\alpha } + \frac{1}{\beta } = \frac{{\alpha + \beta }}{{\alpha \beta }}\\ = \frac{{ - {\rm{ }}p}}{q}\\ = \underline{\underline { - \frac{p}{q}}} \end{array}$$
(ii)
$$\begin{array}{1}\frac{\alpha }{\beta } + \frac{\beta }{\alpha } = \frac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }}\\ = \frac{{({\alpha ^2} + 2\alpha \beta + {\beta ^2}) - 2\alpha \beta }}{{\alpha \beta }}\\ = \frac{{{{(\alpha + \beta )}^2}}}{{\alpha \beta }} - 2\\ = \frac{{{{( - {\rm{ }}p)}^2}}}{q} - 2\\ = \underline{\underline {\frac{{{p^2}}}{q} - 2}} \end{array}$$

(b) Sum of roots of the required equation $$\begin{array}{1} = (\frac{2}{\alpha } + \alpha ) + (\beta + \frac{2}{\beta })\\ = (\alpha + \beta ) + 2(\frac{1}{\alpha } + \frac{1}{\beta })\\ = - {\rm{ }}p + 2( - \frac{p}{q})\\ = - {\rm{ }}p - \frac{{2p}}{q}\end{array}$$

Product of roots of the required equation $$\begin{array}{l} = (\frac{2}{\alpha } + \alpha )(\frac{2}{\beta } + \beta )\\ = \frac{4}{{\alpha \beta }} + \frac{{2\beta }}{\alpha } + \frac{{2\alpha }}{\beta } + \alpha \beta \\ = \frac{4}{{\alpha \beta }} + \alpha \beta + 2(\frac{\alpha }{\beta } + \frac{\beta }{\alpha })\\ = \frac{4}{q} + q + 2(\frac{{{p^2}}}{q} - 2)\\ = \frac{{2{p^2}}}{q} + \frac{4}{q} + q - 4\end{array}$$
∴ A required equation is
$$\begin{array}{1}{x^2} - ( - p - \frac{{2p}}{q})x + (\frac{{2{p^2}}}{q} + \frac{4}{q} + q - 4) = 0\\q{x^2} + p(q + 2)x + (2{p^2} + {q^2} - 4q + 4) = 0\end{array}$$