Question

How do you find the real or imaginary solutions of the equation $ {x^3} + 64 = 0 $ ?

Answer 1

Hint: We are given an algebraic expression as the constants and alphabets are connected to each other by arithmetic operations. We know that when the unknown variable quantity in an algebraic expression is raised to some non-negative integer as the power, we get a polynomial equation and the highest power of the unknown quantity in the equation is known as the degree of the polynomial equation. The given equation has a degree of 3 as the highest exponent of x in the given equation is 3, so the polynomial equation is called a cubic equation. We know that a polynomial equation has exactly as many zeros as the degree of the equation so the given equation has 3 zeros.

Complete step by step solution:
We have to find the solutions of the equation $ {x^3} + 64 = 0 $
It can also be written as $ {x^3} + {(4)^3} = 0 $
We know that –
 $
  {a^3} + {b^3} = (a + b)({a^2} - ab + {b^2}) \\
   \Rightarrow {x^3} + {(4)^3} = (x + 4)({x^2} - 4x + 16) \;
  $
So, we get –
 $
  (x + 4)({x^2} - 4x + 16) = 0 \\
   \Rightarrow x + 4 = 0,\,{x^2} - 4x + 16 = 0 \\
   \Rightarrow x = - 4,\,{x^2} - 4x + 16 = 0 \;
  $
 $ {x^2} - 4x + 16 = 0 $ has a degree 2, so it is a quadratic equation. We can find its solutions by using the quadratic formula –
 $
  x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
  x = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4(1)(16)} }}{{2(1)}} \\
   \Rightarrow x = \dfrac{{4 \pm \sqrt {16 - 64} }}{2} \\
   \Rightarrow x = \dfrac{{4 \pm \sqrt { - 48} }}{2} \;
  $
We know that $ \sqrt { - 1} = i $ so $ \sqrt { - 48} = \sqrt {48} i $
 $
   \Rightarrow x = \dfrac{{4 \pm 4\sqrt 3 i}}{2} \\
   \Rightarrow x = 2 \pm 2\sqrt 3 i \;
  $
Hence the solutions of the equation $ {x^3} + 64 = 0 $ are $ x = - 4,\,x = 2 + 2\sqrt 3 i $ and $ x = 2 - 2\sqrt 3 i $ .
So, the correct answer is “$ x = - 4,\,x = 2 + 2\sqrt 3 i $ and $ x = 2 - 2\sqrt 3 i $ ”.

Note: The obtained quadratic equation could not be solved by factorization so we used the quadratic formula to find the values of x. The solution $ x = - 4 $ is a real solution as it is an integer that can be shown on the number line while the solutions $ x = 2 + 2\sqrt 3 i $ and $ x = 2 - 2\sqrt 3 i $ are imaginary as they involve iota.