# 10 Deadly Common Algebra Mistakes

# Common Patterns of Algebra Mistakes

Students often make Common Algebra Mistakes due to confusion, such as expanding and simplifying 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 whether you make similar mistakes because, surprisingly, even university students still make 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!

The 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.

The 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

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

The correct solution would be;

\( (x-3)^2 = x^2-6x + 9 \)

### Pattern 5

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

The correct solution would be;

\( (x+3)^2 = x^2 +6x + 9 \)

### Pattern 6

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

The correct solution would be;

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

### Pattern 7

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

The coefficient and the base cannot be multiplied.

The 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.

The correct solution would be;

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

### Pattern 9

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

The 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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