How do you solve #27x^3 + 343 = 0#?

2 Answers
Sep 18, 2016

Answer:

#x=-7/3#

Explanation:

#27x^3+343" is a sum of cubes"# and factorises, in general, as follows.

#color(red)(bar(ul(|color(white)(a/a)color(black)(a^3+b^3=(a+b)(a^2-ab+b^2))color(white)(a/a)|)))#

#(3x)^3=27x^3" and " (7)^3=343rArra=3x" and " b=7#

#rArr27x^3+343=(3x+7)(9x^2-21x+49)#

#(3x+7)(9x^2-21x+49)=0#

solve : #3x+7=0rArrx=-7/3#

solve : #9x^2-21x+49=0#

Before attempting to factorise check the #color(blue)"discriminant"#

here a = 9 , b = - 21 and c = 49

#b^2-4ac=(-21)^2-(4xx9xx49)=-1323<0#

Hence this quadratic has no real roots.

The only real solution is therefore #x=-7/3#

Sep 18, 2016

Answer:

#x = -7/3#

Explanation:

It is a real advantage to know all the powers up to 1000 by heart!

In such a case, we could go the conventional method of factoring and solving fo each factor equal to 0.

However, if you recognise that both 27 and 343 are cubes, as there is only one variable, we can also write as follows:

#27x^3 +343 = 0#

#27x^3 = -343#

#x^3 = -343/27#

#x = root3(-343/27)#

It is possible to have a cube root of a negative value.

#x = -7/3#