Question

How do you solve $4{x^2} + 4ax + \left( {{a^2} + {b^2}} \right) = 0$ using the quadratic formula?

Answer 1

Hint: The given problem requires us to solve a quadratic equation using quadratic formula. There are various methods that can be employed to solve a quadratic equation like completing the square method, using quadratic formulas and by splitting the middle term. Using the quadratic formula gives us the roots of the equation directly with ease.

Complete step-by-step answer:
In the given question, we are required to solve the equation $4{x^2} + 4ax + \left( {{a^2} + {b^2}} \right) = 0$ with the help of quadratic formula. The quadratic formula can be employed for solving an equation only if we compare the given equation with the standard form of the quadratic equation.
Comparing with standard quadratic equation $4{x^2} + 4ax + \left( {{a^2} + {b^2}} \right) = 0$
Here,$a = 4$, $b = 4a$ and$c = {a^2} + {b^2}$.
Now, Using the quadratic formula, we get the roots of the equation $A{x^2} + Bx + C = 0$ as:
$x = \dfrac{{ - B \pm \sqrt {{B^2} - 4AC} }}{{2A}}$
Substituting the values of A, B, and C in the quadratic formula, we get,
$ \Rightarrow x = \dfrac{{ - \left( {4a} \right) \pm \sqrt {{{\left( {4a} \right)}^2} - 4 \times 4 \times \left( {{a^2} + {b^2}} \right)} }}{{2 \times 4}}$
Putting the values of
$ \Rightarrow x = \dfrac{{ - 4a \pm \sqrt {16{a^2} - 16\left( {{a^2} + {b^2}} \right)} }}{{2 \times 4}}$
$ \Rightarrow x = \dfrac{{ - 4a \pm \sqrt {16{a^2} - 16{a^2} - 16{b^2}} }}{{2 \times 4}}$
Cancelling the like terms with opposite signs, we get,
$ \Rightarrow x = \dfrac{{ - 4a \pm \sqrt { - 16{b^2}} }}{{2 \times 4}}$
Now, we know that square of $4b$ is $16{b^2}$. So, we get,
$ \Rightarrow x = \dfrac{{ - 4a \pm \sqrt { - {{\left( {4b} \right)}^2}} }}{{2 \times 4}}$
So, taking the perfect square root term outside of the square root, we get,
$ \Rightarrow x = \dfrac{{ - 4a \pm \left( {4b} \right)\sqrt { - 1} }}{{2 \times 4}}$
Now, we know that the square root of $\left( { - 1} \right)$ is iota, i.
$ \Rightarrow x = \dfrac{{ - 4a \pm 4bi}}{8}$
Separating the denominator, we get,
$ \Rightarrow x = \dfrac{{ - 4a}}{8} \pm \dfrac{{4bi}}{8}$
Cancelling the common factors in numerator and denominator, we get,
$ \Rightarrow x = - \left( {\dfrac{a}{2}} \right) \pm i\left( {\dfrac{b}{2}} \right)$
So, \[x = - \left( {\dfrac{a}{2}} \right) + i\left( {\dfrac{b}{2}} \right)\] and \[x = - \left( {\dfrac{a}{2}} \right) - i\left( {\dfrac{b}{2}} \right)\] are the roots of the equation $4{x^2} + 4ax + \left( {{a^2} + {b^2}} \right) = 0$.
So, the roots of the equation $4{x^2} + 4ax + \left( {{a^2} + {b^2}} \right) = 0$ are: \[x = - \left( {\dfrac{a}{2}} \right) + i\left( {\dfrac{b}{2}} \right)\] and \[x = - \left( {\dfrac{a}{2}} \right) - i\left( {\dfrac{b}{2}} \right)\] .
So, the correct answer is “ \[x = - \left( {\dfrac{a}{2}} \right) + i\left( {\dfrac{b}{2}} \right)\] and \[x = - \left( {\dfrac{a}{2}} \right) - i\left( {\dfrac{b}{2}} \right)\] .”.

Note: Quadratic equations are the polynomial equations with degree of the variable or unknown as 2. Quadratic formula is the easiest and most efficient formula to calculate the roots of an equation. Quadratic equations can also be solved by splitting the middle term and completing the square method. Quadratic equations may also be solved by hit and trial method if the roots of the equation are easy to find.