Algebra

First minus 49 both sides

\(x^2 + 10x - 24 = 0\)

Using the Quadratic cross method ( https://www.youtube.com/watch?v=at7MIBm-it8&ab_channel=MisterBrash )

\((x-2)(x+12) = 0\)

So \(x\) have to be either 2 or -12. And we are finding the smallest value so it's -12

 

Find the smallest value of x such that  x^2 + 10x + 25 = 49.    

 

                                                 x² + 10x + 25  =  49 

 

Subtract 49 from both sides      x² + 10x – 24  =  0  

 

Factor left side                          (x + 12)(x – 2)  =  0  

 

Set each factor to zero                        (x + 12)  =  0   

                                                            (x – 2)  =  0   

 

So    x = –12  and  x = +2    The smaller of these is x = –12  

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