Solution assignment 08 The basics

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Assignment 8

Solve the following equations:

(x-3)^2+2(x-3)=0

(x-5)^2-16=0

x^2+x-1=0

Solution

We can solve the first equation by taking x-3 outside brackets. We get:

(x-3)[(x-3)+2]=0, so:

(x-3)(x-1)=0, dus:

x=3 of x=1

We can write the second equation as:

x-5=4 of x-5=-4, so:

x=9 of x=1

The last equation cannot be solved by factorizion. Then the abc-formula has to be used and always delivers a solution if it exists. We get:

x=\displaystyle\frac{-1\pm\sqrt{1+4}}{2}, and thus:

x=-\displaystyle\frac{1}{2}+\frac{1}{2}\sqrt{5} of x=-\displaystyle\frac{1}{2}-\frac{1}{2}\sqrt{5}

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