Question #266570

Examine the following steps in the solution to 5π‘₯2 βˆ’ 5 = 0

5(π‘₯2 βˆ’ 1) = 0

5(π‘₯ βˆ’ 1)(π‘₯ + 1) = 0

π‘₯ βˆ’ 1 = 0

or

π‘₯ + 1 = 0

π‘₯ = 1

or

π‘₯

= βˆ’1

What happened to 5? Explain.


Expert's answer

5x2βˆ’5=05x^2βˆ’5=0

This is the given equation

5x2βˆ’5=05(x2βˆ’1)=0(x2βˆ’1)=05x2βˆ’1=0(x+1)(xβˆ’1)=0x=1orx=βˆ’15x^2βˆ’5=0\\5(x^2βˆ’1)=0\\(x^2βˆ’1)=\frac{0}{5}\\x^2βˆ’1=0\\(x+1)(xβˆ’1)=0\\x=1 \\or\\ x=βˆ’1


5 got divided by 0 



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