Quadratic Equations in One Unknown-Examples


  • 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.
  • Solutions
    (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.


  • Examples
    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) \((\alpha + 3)(\beta - 3)\)
  • Solutions
    \(\begin{array}{c}x(2x + 3) = 1\\2{x^2} + 3x - 1 = 0\end{array}\)
    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)
    \(\begin{array}{1}(\alpha + 3)(\beta - 3) = \alpha \beta - 3(\alpha - \beta ) - 9\\ = - \frac{1}{2} - 3\sqrt {\frac{{17}}{4}} - {9^{^{^{^{^{}}}}}}\\ = {\underline{\underline {\frac{{ - {\rm{ }}19 - 3\sqrt {17} }}{2}}} ^{}}\end{array}\)


  • Examples
    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 \((\beta + \frac{2}{\beta })\).
  • Solutions
    (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
    \(\begin{array}{1}{x^2} - ( - p - \frac{{2p}}{q})x + (\frac{{2{p^2}}}{q} + \frac{4}{q} + q - 4) = 0\\q{x^2} + p(q + 2)x + (2{p^2} + {q^2} - 4q + 4) = 0\end{array}\)