Introduction
This quadratic formula equation is a strong Secondary Maths example because students must first clear fractions carefully, then solve the resulting quadratic correctly. Many students rush the algebra and make sign mistakes, but once the equation is rearranged properly, this quadratic formula equation becomes much more systematic.
The Question / Scenario Explanation
Source: Solve by Quadratic Formula (2) Sec 3 G2 and G3
Question (as shown):
\(\frac{x}{x-1}-\frac{2}{2-3x}=0.1\)
This question asks students to solve the equation using the quadratic formula.
Step-by-Step Solution / Explanation
Step 1: Note the denominator restrictions
Before solving this quadratic formula equation, check the denominators:
\(x-1 \neq 0 \Rightarrow x \neq 1\)
\(2-3x \neq 0 \Rightarrow x \neq \frac{2}{3}\)
These values cannot be part of the final answer.
Step 2: Clear the fractions
Multiply both sides by \((x-1)(2-3x)\):
\(\frac{x}{x-1}(x-1)(2-3x)-\frac{2}{2-3x}(x-1)(2-3x)=0.1(x-1)(2-3x)\)
This simplifies to:
\(x(2-3x)-2(x-1)=0.1(x-1)(2-3x)\)
Step 3: Expand both sides
Expand the left-hand side:
\(x(2-3x)-2(x-1)=2x-3x^2-2x+2=-3x^2+2\)
Expand the right-hand side:
\((x-1)(2-3x)=2x-3x^2-2+3x=-3x^2+5x-2\)
So:
\(0.1(x-1)(2-3x)=-0.3x^2+0.5x-0.2\)
The equation becomes:
\(-3x^2+2=-0.3x^2+0.5x-0.2\)
Step 4: Rearrange into quadratic form
Bring everything to one side:
\(-3x^2+2+0.3x^2-0.5x+0.2=0\)
\(-2.7x^2-0.5x+2.2=0\)
To remove decimals, multiply the whole equation by \(-10\):
\(27x^2+5x-22=0\)
This is now in the standard quadratic form for the quadratic formula equation.
Step 5: Use the quadratic formula
For \(27x^2+5x-22=0\), we have:
\(a=27,\ b=5,\ c=-22\)
Use the formula:
\(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
Substitute the values:
\(x=\frac{-5\pm\sqrt{5^2-4(27)(-22)}}{2(27)}\)
\(x=\frac{-5\pm\sqrt{25+2376}}{54}\)
\(x=\frac{-5\pm\sqrt{2401}}{54}\)
\(x=\frac{-5\pm49}{54}\)
Step 6: Find the two solutions
First solution:
\(x=\frac{-5+49}{54}=\frac{44}{54}=\frac{22}{27}\)
Second solution:
\(x=\frac{-5-49}{54}=\frac{-54}{54}=-1\)
So the two solutions are:
\(x=\frac{22}{27}\) or \(x=-1\)
Both values are valid because neither is \(1\) nor \(\frac{2}{3}\).
✅ Final Answer: \(x=\frac{22}{27}\) or \(x=-1\)
Key Concepts Students Must Know
- In a quadratic formula equation, fractions often need to be cleared before the quadratic formula can be used.
- Always check denominator restrictions first so invalid values are not accepted later.
- If decimals appear after expansion, multiply through by a suitable number to make the equation easier to solve.
- The quadratic formula is useful when factorisation is not obvious at the beginning.
- After solving, students should still check whether the answers are allowed by the original equation.
Exam Tips / Common Mistakes
Exam Tips
- Write the denominator restrictions before doing any algebra.
- For this quadratic formula equation, clear the fractions carefully using the full product of both denominators.
- Expand one bracket at a time to avoid sign mistakes.
- Remove decimals before using the quadratic formula if it makes the equation cleaner.
- Check that your final answers do not make any denominator equal to zero.
Common Mistakes
- Forgetting that \(\frac{2}{2-3x}\) has a negative sign in the denominator expression when expanding.
- Expanding \((x-1)(2-3x)\) wrongly.
- Using the quadratic formula on the decimal version and making calculator or sign errors.
- Getting correct roots but forgetting to check whether \(x=1\) or \(x=\frac{2}{3}\) must be excluded.
- Stopping at \(\frac{-5\pm49}{54}\) without simplifying fully.
Parent Insight
This quadratic formula equation is a good example of how Secondary Maths combines algebraic manipulation with method selection. Many students know the quadratic formula, but still lose marks because they do not clear the fractions carefully first. With guided practice, students become more confident in turning complicated-looking equations into clean quadratics they can solve accurately.
Conclusion
To solve this quadratic formula equation, we first cleared the fractions, expanded both sides, and simplified the expression to \(27x^2+5x-22=0\). Then we applied the quadratic formula and obtained \(x=\frac{22}{27}\) or \(x=-1\). Both answers are valid because they do not break the original denominator restrictions.
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