Question

How many solutions are there to \[{x^2} + 3x + 5 = 0\] ?

Answer 1

Hint: Since for any quadratic equation, the solution means those values for the variable that give the value of the equation as $0$, then find roots of the equation. Use a method of determinant to solve for the value of $x$ from the given quadratic equation. Compare the quadratic equation with general quadratic equation and substitute values in the formula of finding roots of the equation. Solve the value under the square root and write two zeros by separating plus and minus signs from the obtained answer.

For a general quadratic equation \[a{x^2} + bx + c = 0\], roots are given by formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Complete step by step answer:
We are given the quadratic equation \[{x^2} + 3x + 5 = 0\] ---- (1)
We know that the general quadratic equation is \[a{x^2} + bx + c = 0\] where ‘a’, ‘b’, and ‘c’ are constant values.
On comparing the quadratic equation in equation (1) with general quadratic equation\[a{x^2} + bx + c = 0\], we get \[a = 1,b = 3,c = 5\]
Substitute the values of a, b and c in the formula of finding roots of the equation.
\[ \Rightarrow x = \dfrac{{ - (3) \pm \sqrt {{{(3)}^2} - 4 \times 1 \times 5} }}{{2 \times 1}}\]
Square the terms under the square root in numerator of the fraction
\[ \Rightarrow x = \dfrac{{ - 3 \pm \sqrt {9 - 20} }}{2}\]
Calculate the subtraction inside the square root
\[ \Rightarrow x = \dfrac{{ - 3 \pm \sqrt { - 11} }}{2}\]
We can write \[\sqrt { - 11} = \sqrt {11} i\]. Substitute this value in the numerator of the fraction.
\[ \Rightarrow x = \dfrac{{ - 3 \pm i\sqrt {11} }}{2}\]
Roots are \[\dfrac{{ - 3 + i\sqrt {11} }}{2}\] and \[\dfrac{{ - 3 - i\sqrt {11} }}{2}\]
So, there are no real roots for the quadratic equation \[{x^2} + 3x + 5 = 0\]
The only roots are complex roots for the equation.
\[\therefore \]The solution for the equation \[{x^2} + 3x + 5 = 0\] are \[\dfrac{{ - 3 + i\sqrt {11} }}{2}\] and \[\dfrac{{ - 3 - i\sqrt {11} }}{2}\]

Note: Do not leave their solution incomplete after they get complex values for the root. Keep in mind the solutions of the equation are the values of $x$ that make the value of the equation equal to $0$.