10 Deadly Common Algebra Mistakes

Common Patterns of Algebra Mistakes

Students often make Common Algebra Mistakes due to confusions such as expand and simplify rules, fractions, indices and equations which that lead the students to the wrong answer. Also, these error patterns are very basic and quite easily rectified. Check yourself whether you make similar mistakes because surprisingly even students in the university are still making these silly mistakes.
Let me start with very simple patterns.

Pattern 1

\( \displaystyle \frac{4}{9} = \frac{2}{3} \)

Many of students may think this is OK, but \( \displaystyle \frac{4}{9} \ne \frac{2}{3} \). Do you think so?
\( \displaystyle \sqrt{\frac{4}{9}} = \frac{2}{3} \) is TRUE!

Pattern 2

\( \begin{aligned} \displaystyle
x^2 &= 16 \\
x &= 4 \\
\end{aligned} \)

Quite often my students made this mistake in their school exams. Bad!
Correct solution would be;
\( \begin{aligned} \displaystyle
x^2 &= 16 \\
x &= \pm 4
\end{aligned} \)

Pattern 3

\( \begin{aligned} \displaystyle
x^2 &= 2x \\
x &= 2 \\
\end{aligned} \)

What do you think? You cannot cancel \( x \) from both sides.
Correct solution would be;
\( \begin{aligned} \displaystyle
x^2 &= 2x \\
x – 2x &= 0 \\
x(x-2) &= 0 \\
x &= 0 \text{ or } x = 2
\end{aligned} \)

Pattern 4

\( (x-3)^2 = x^2 -9 \)

Correct solution would be;
\( (x-3)^2 = x^2 -6x + 9 \)

Pattern 5

\( (x+3)^2 = x^2 +9 \)

Correct solution would be;
\( (x+3)^2 = x^2 +6x + 9 \)

Pattern 6

\( \sqrt{x^2 + y^2} = x + y \)

Correct solution would be;
\( \sqrt{(x+y)^2} = x + y \)

Pattern 7

\( 3 \times 2^x = 6^x \)

Coefficient and the base cannot be multiplied.
Correct solution would be;
\( 3^x \times 2^x = 6^x \)

Pattern 8

\( \sqrt{x^2} = x \)

The outcome of the square root must be positive.
Correct solution would be;
\( \sqrt{x^2} = |x| \)

Pattern 9

\( \displaystyle a-\frac{b-c}{d} = \frac{ad-b-c}{d} \)

Correct solution would be;
\( \begin{aligned} \displaystyle
a-\frac{b-c}{d} &= \frac{ad-(b-c)}{d} \\
&= \frac{ad-b+c}{d}
\end{aligned} \\ \)

Pattern 10

\( \begin{aligned} \displaystyle
\frac{x-2}{2} &= \frac{x – 1}{1} \\
&= x-1
\end{aligned} \)

You cannot cancel 2 at the numerator and 2 at the denominator.

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