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Question: What are the solutions to the equation \({x^2} + 2x + 2 = 0\)?...

What are the solutions to the equation x2+2x+2=0{x^2} + 2x + 2 = 0?

Explanation

Solution

The given equation is a quadratic equation so it will have two roots. To find these roots, we are going to use the quadratic formula, that is
⇒x=−b±b2−4ac2a\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
By putting the values of a,b,ca,b,c we will get the roots of our equation.

Complete step-by-step solution:
In this question, we have to find the solution of the equation x2+2x+2=0{x^2} + 2x + 2 = 0.
Now, this is a quadratic equation and it will have two possible answers.
For solving this quadratic equation, we are going to use the quadratic formula.
The quadratic formula for a given quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0 is given below:
⇒x=−b±b2−4ac2a\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} - - - - - - - - - - - - - - (1)
Where, bb is the coefficient of xxand aa is the coefficient of x2{x^2}and c is the constant.
Our given equation is: x2+2x+2=0{x^2} + 2x + 2 = 0
Therefore, a=1,b=2,c=2a = 1,b = 2,c = 2
Putting this values in equation (1), we get
⇒x=−2±22−4×1×22×1\Rightarrow x = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4 \times 1 \times 2} }}{{2 \times 1}}
Solving the equation, we get
⇒x=−2±4−82 ⇒x=−2±−42  \Rightarrow x = \dfrac{{ - 2 \pm \sqrt {4 - 8} }}{2} \\\ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt { - 4} }}{2} \\\
Now, since there is a negative number under the square root, the roots of the equation will be imaginary.
And we know that we can denote imaginary numbers with complex numbers ii.
Therefore, above equation becomes
⇒x=−2±2i2 ⇒x=−2+2i2 ⇒x=2(i−1)2 ⇒x=i−1  \Rightarrow x = \dfrac{{ - 2 \pm 2i}}{2} \\\ \Rightarrow x = \dfrac{{ - 2 + 2i}}{2} \\\ \Rightarrow x = \dfrac{{2\left( {i - 1} \right)}}{2} \\\ \Rightarrow x = i - 1 \\\ ⇒x=−2±2i2 ⇒x=−2−2i2 ⇒x=2(−i−1)2 ⇒x=−i−1  \Rightarrow x = \dfrac{{ - 2 \pm 2i}}{2} \\\ \Rightarrow x = \dfrac{{ - 2 - 2i}}{2} \\\ \Rightarrow x = \dfrac{{2\left( { - i - 1} \right)}}{2} \\\ \Rightarrow x = - i - 1 \\\
Therefore, the imaginary values of xx in equation x2+2x+2=0{x^2} + 2x + 2 = 0 are x=i−1x = i - 1 and x=−i−1x = - i - 1.

Note: In the quadratic formula,
⇒x=−b±b2−4ac2a\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
The term b2−4ac{b^2} - 4ac is known as the discriminant(D)\left( D \right) of the equation.
(i). If the value of D>0D > 0, then the roots of the equation are real and distinct.
(ii). If the value of D=0D = 0, then the equation has only one root.
(iii). If the value of D<0D < 0, then the equation has imaginary roots.