Question

Show that the equation $ {x^2} + ax - 4 = 0 $ has real and distinct roots for all real values of a.

Answer 1

Hint: To discuss the nature of roots of a given quadratic equation. We first find or write all its values of a, b and c from a given equation and then we use these values in the discriminant formula of the quadratic equation to see whether its result is negative, zero or positive. If its result is positive then we can say roots are distinct and real.
Formulas used: Discriminant of a quadratic equation: $ D = {b^2} - 4ac $

Complete step-by-step answer:
Given equation is $ {x^2} + ax - 4 = 0 $
Comparing above equation with standard quadratic equation. Which is $ a{x^2} + bx + c = 0 $
Therefore, we have:
 $ a = 1,\,\,b = \,\,a\,\,and\,\,c = - 4 $
Also, we know that to discuss whether a given quadratic equation has real and distinct roots we always calculate the discriminant of the quadratic equation and if its value comes out to be either equal to zero or greater than zero then we can say that roots are real and distinct according to the value we will get.
Now, calculating discriminant of a given quadratic equation.
Discriminant is given as:
 $ D = {b^2} - 4ac $
Substituting values in above formula. We have,
 $
  D = {\left( a \right)^2} - 4(1)\left( { - 4} \right) \\
   \Rightarrow D = {a^2} + 16 \;
  $
From above we clearly see that the value of D will never be negative for any value of ‘a’.
Therefore, $ D > 0 $ for all values of ‘a’.
Hence, we can say that given a quadratic equation $ {x^2} + ax - 4 = 0 $ will have real and distinct roots for all real values of ‘a’.

Note: To find whether a given equation will have real roots or not. We always calculate discriminant of given equation and its value will decide the nature of given equation as we know that for:
 $ D = 0 $ , we have real and equal roots.
 $ D > 0 $ , roots are real but distinct or we can say unequal
 $ D < 0 $ , roots are not real or we can say in this case roots will be complex roots.