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2013 National Preliminary mathematics Topic 29 Free

Order of operations and numeric expressions

Inserting parentheses to make $3+5^2-8\times3+5=0$ true · Sub-topic 1

PRELIMINARY STAGE

2013

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QUESTION

Insert parentheses in the equation so that it is true: $3+5^2-8\times3+5=0$.

ANSWER: $(3+5)^2-8\times(3+5)=0$

SOLUTION

$(3+5)^2-8(3+5)=64-64=0$.


PRACTICE QUESTIONS

Insert parentheses so that each equation is true.


1. $2+4^2-6\times2+4=0$

ANSWER: $(2+4)^2-6\times(2+4)=0$


SOLUTION

Group $2+4$ so that it is squared and also multiplied by $6$:

$(2+4)^2-6\times(2+4)=36-36=0$


2. $1+3^2-4\times1+3=0$

ANSWER: $(1+3)^2-4\times(1+3)=0$


SOLUTION

Group $1+3$ so that it is squared and also multiplied by $4$:

$(1+3)^2-4\times(1+3)=16-16=0$


3. $4+2^2-6\times4+2=0$

ANSWER: $(4+2)^2-6\times(4+2)=0$


SOLUTION

Group $4+2$ so that it is squared and also multiplied by $6$:

$(4+2)^2-6\times(4+2)=36-36=0$


4. $3+6^2-9\times3+6=0$

ANSWER: $(3+6)^2-9\times(3+6)=0$


SOLUTION

Group $3+6$ so that it is squared and also multiplied by $9$:

$(3+6)^2-9\times(3+6)=81-81=0$


5. $2+7^2-9\times2+7=0$

ANSWER: $(2+7)^2-9\times(2+7)=0$


SOLUTION

Group $2+7$ so that it is squared and also multiplied by $9$:

$(2+7)^2-9\times(2+7)=81-81=0$


6. $5+1^2-6\times5+1=0$

ANSWER: $(5+1)^2-6\times(5+1)=0$


SOLUTION

Group $5+1$ so that it is squared and also multiplied by $6$:

$(5+1)^2-6\times(5+1)=36-36=0$