Question

How do you use a quadratic formula to solve $2{{\text{x}}^2} + 4{\text{x}} + 2 = 0$?

Answer 1

Hint: In this question, we are asked to solve the given equation using the quadratic equation formula. First we have to check the existences of roots using the formula stated below and find the types of roots.
After that we have to apply the quadratic equation formula and derive the answer.

Formula used: Quadratic equation formula:
${\text{x = }}\dfrac{{{{ - b \pm }}\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} }}{{{\text{2a}}}}$
To check the roots = $\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} $

Complete step-by-step solution:
The given equation is $2{{\text{x}}^2} + 4{\text{x}} + 2 = 0$, we need to solve the equation using the quadratic equation formula.
Quadratic equation formula:
 ${\text{x = }}\dfrac{{{{ - b \pm }}\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} }}{{{\text{2a}}}}$
Before that, we have to check the roots. For that we will use this formula $\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} $
Clearly in $2{{\text{x}}^2} + 4{\text{x}} + 2 = 0$,
${\text{a = 2}}$,
${\text{b = 4}}$,
${\text{c = 2}}$.
$\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} = \sqrt {{\text{16 - 16}}} $
$\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} = 0$
Therefore, the roots exist and the roots are equal.
Now applying $2{{\text{x}}^2} + 4{\text{x}} + 2 = 0$ in $\dfrac{{{{ - b \pm }}\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} }}{{{\text{2a}}}}$,
${\text{x = }}\dfrac{{ - 4 \pm \sqrt {{4^2} - 4(2)(2)} }}{{2(2)}}$
On simplification we get
${\text{x = }}\dfrac{{ - 4 \pm \sqrt {16 - 4(4)} }}{4}$
Simplifying the numerator we get,
${\text{x = }}\dfrac{{ - 4 \pm \sqrt 0 }}{4}$
Now we can expand the expression into two, as there is a $ \pm $ in the expression. One becomes plus and the other becomes minus.
${\text{x = }}\dfrac{{ - 4 + 0}}{4},\dfrac{{ - 4 - 0}}{4}$
On adding the term and we get
${\text{x = }}\dfrac{{ - 4}}{4},\dfrac{{ - 4}}{4}$
Now we get the value of ${\text{x}}$ which are equal roots.
\[{\text{x = }} - 1, - 1\]

Therefore the root of $2{{\text{x}}^2} + 4{\text{x}} + 2 = 0$ is \[{\text{ }} - 1\].

Note: In this type of questions, students always miss an important step that will reduce their marks. Whenever the question asks you to find the roots with a quadratic equation, first you need to check the roots whether the roots exist or not and if it exists is it clear roots or equal roots. This is an important step.
 To check the roots, we have to find the value of $\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} $ first,
If $\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} $ is greater than $0$ , then the root are different and clear.
If $\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} $ is equal to $0$, then the roots are equal.
If $\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} $ is less that $0$, then the roots does not exist