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Functions, 2ADV EQ-Bank 28

Given \(p\) and \(q\) are rational numbers, and  \(p, q \neq 0\), show

\(px^2-(p+q) x+q=0\)

has rational roots.   (3 marks)

--- 10 WORK AREA LINES (style=lined) ---

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\(\text{Proof (See Worked Solution)}\)

Show Worked Solution
\(\Delta\) \(=b^2-4 a c\)
  \(=[-(p+q)]^2-4 \times p \times q\)
  \(=p^2+2 p q+q^2-4 p q\)
  \(=p^2-2 p q+q^2\)
  \(=(p-q)^2\)

 

\(\text{Roots of equation using quadratic formula:}\)

\(x\) \(=\dfrac{(p+q) \pm \sqrt{(p-q)^2}}{2 p}\)
  \(=\dfrac{p+q+(p-q)}{2 p} \ \ \text{or} \ \ \dfrac{p+q-(p-q)}{2 p}\)
  \(=1 \ \ \text{or} \ \ \dfrac{q}{p}\).

 

\(\text{Since \(p, q\) are rational, all roots are rational.}\)

Filed Under: Quadratics and Cubic Functions Tagged With: Band 5, smc-6215-80-Discriminant

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