Question

How do you factor and solve $ {t^2} + 16 = 0 $ ?

Answer 1

Hint: In order to determine the solution to the above quadratic equation having variable $ x $ ,first we compare it with the standard equation i.e. $ a{x^2} + bx + c $ to obtain the value of the coefficients. Then find the determinant to find out the nature of roots . If the root comes to be $ D > 0 $ which means roots are distinct and real and in our case it comes out to be $ D < 0 $ which means the roots are imaginary . Thereafter , now apply the quadratic formula $ x = \dfrac{{ - b \pm \sqrt D }}{{2a}} $ to get your required solutions.

Complete step-by-step answer:
Given a quadratic equation, $ {t^2} + 16 = 0 $ let it be $ f(x) $ and the roots to the equation be $ {x_1},{x_2} $ .
 $ f(x) = 1{t^2} - 0t + 16 = 0 $
Comparing the equation with the standard Quadratic equation $ a{x^2} + bx + c $
a becomes 1
b becomes 0
And c becomes 16
Let’s now find the determinant value of this quadratic equation, as by looking at the value of determinant we can find out the nature of both the roots .
 $ D = {b^2} - 4ac $
Putting the values, we get
 $
  D = {\left( 0 \right)^2} - 4(1)(16) \\
   = - 64 \;
  $
As we can see $ D > 0 $ ,so we can conclude that the both roots of the equation will be distinct and real.
Now using Quadratic formula to find out the actual value of the roots
x= $ \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
Since, $ {b^2} - 4ac = D $
 $
  x = \dfrac{{ - b \pm \sqrt D }}{{2a}} \\
   \Rightarrow {x_1} = \dfrac{{ - b + \sqrt D }}{{2a}},{x_2} = \dfrac{{ - b - \sqrt D }}{{2a}} \;
  $
Now putting the value of $ D,b,a $ ,we get
 $
   \Rightarrow {x_1} = \dfrac{{ - (0) + \sqrt { - 64} }}{{2(1)}},{x_2} = \dfrac{{ - (0) - \sqrt { - 64} }}{{2(1)}} \\
    \\
   \Rightarrow {x_1} = \dfrac{{ + \sqrt { - 64} }}{2},{x_2} = \dfrac{{ - \sqrt { - 64} }}{2} \;
  $
Simplifying more further, $ \sqrt { - 64} = 8i $
 $
   \Rightarrow {x_1} = \dfrac{{ + 8i}}{2},{x_2} = \dfrac{{ - 8i}}{2} \\
  i = \sqrt { - 1} \;
  $
Therefore , the solution to the quadratic equation $ {t^2} + 16 = 0 $ having imaginary roots $ {x_1},{x_2} $ is $
   \Rightarrow {x_1} = \dfrac{{ + 8i}}{2},{x_2} = \dfrac{{ - 8i}}{2} \\
   \;
  $
So, the correct answer is “Option C”.

Note: Quadratic Equation: A quadratic equation is a equation which can be represented in the form of $ a{x^2} + bx + c $ where $ x $ is the unknown variable and a,b,c are the numbers known where $ a \ne 0 $ .If $ a = 0 $ then the equation will become linear equation and will no more quadratic .
The degree of the quadratic equation is of the order 2.
Every Quadratic equation has 2 roots.
Discriminant: $ D = {b^2} - 4ac $
Using Discriminant, we can find out the nature of the roots
If D is equal to zero, then both of the roots will be the same and real.
If D is a positive number then, both of the roots are real solutions.
If D is a negative number, then the root are the pair of complex solutions