105th Problem 2016

Algebra Level 1

Solve the quadratic equation:

x 2 + 14 x + 48 = 0 { x }^{ 2 }+14x+48=0

Check out the set: 2016 Problems .
x = − 6 or x = − 8 x=-6 \text{ or } x=-8 x = 6 or x = − 8 x= 6 \text{ or } x= -8 x = 6 or x = 8 x= 6 \text{ or } x= 8 x = − 6 or x = 8 x= -6 \text{ or } x= 8

Angela Fajardo by Angela Fajardo , 19, Philippines

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1 solution

Ashish Menon
Apr 24, 2016

x 2 + 14 x + 48 = 0 x 2 + 6 x + 8 x + 48 = 0 x ( x + 6 ) + 8 ( x + 6 ) = 0 ( x + 6 ) ( x + 8 ) = 0 ( x + 6 ) = 0 or ( x + 8 ) = 0 x ∈ { − 6 , − 8 } \begin{aligned} x^2 + 14x + 48 & = 0\\ x^2 + 6x + 8x + 48 & = 0\\ x(x + 6) + 8(x + 6) & = 0\\ (x + 6)(x + 8) & = 0\\ (x + 6) = 0 \ \text{or} \ (x + 8) = 0\\ x \in \boxed{\{-6,-8\}} \end{aligned}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

You cannot say that x = { − 6 , − 8 } x = \{-6, -8\} because { − 6 , − 8 } \{-6, -8\} is a set. You could, for example, replace it with x ∈ { − 6 , − 8 } x \in \{-6, -8\} .
Also, you should replace "or" with "\text{ or }" to make it look nicer.

Jesse Nieminen - 4 years, 11 months ago

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Yeah thanks,this solution was one long ago when I didnt know nice Latex, now I know them perfectly ;)

Ashish Menon - 4 years, 11 months ago

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Great! Now that this is correct, I'll upvote this.

Jesse Nieminen - 4 years, 11 months ago

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@Jesse Nieminen – :P T h a n k s ! ! \color{#CEBB00}{\mathcal{Thanks!!}} .

Ashish Menon - 4 years, 11 months ago

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