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Question: How do you solve \({x^2} + 6x + 10 = 0\) by completing the square?...

How do you solve x2+6x+10=0{x^2} + 6x + 10 = 0 by completing the square?

Explanation

Solution

Here, we are required to solve the given quadratic equation using the method of completing the square. Hence, we will simply add and subtract the square of a constant and write the term containing xx such that it forms a term in a form 2ab2ab, thus, this will help us to apply the square identities and thus, ‘complete the square’ and solve it further to find the required roots of the given quadratic equation.

Formula Used: (a+b)2=a2+2ab+b2{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}.

Complete step-by-step answer:
Given quadratic equation is:
x2+6x+10=0{x^2} + 6x + 10 = 0
Now, since, we are required to solve this question using completing the square, hence, we will write this quadratic equation as:
(x)2+2(x)(3)+(3)2−(3)2+10=0{\left( x \right)^2} + 2\left( x \right)\left( 3 \right) + {\left( 3 \right)^2} - {\left( 3 \right)^2} + 10 = 0
Hence, even if this quadratic equation was not a perfect square, but we tried to make it a perfect square by adding and subtracting the square of a constant such that, we complete the square by making the identity, (a+b)2=a2+2ab+b2{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}.
Hence, using this identity, we get,
(x+3)2−9+10=0{\left( {x + 3} \right)^2} - 9 + 10 = 0
⇒(x+3)2+1=0\Rightarrow {\left( {x + 3} \right)^2} + 1 = 0
Subtracting 1 from both the sides,
⇒(x+3)2=−1\Rightarrow {\left( {x + 3} \right)^2} = - 1
Now, we know that squares of any number can’t be negative.
Thus, we will use complex numbers or imaginary numbers.
And we know that, −1=±i\sqrt { - 1} = \pm i…………………..(1)\left( 1 \right)
Thus, taking square root on both sides of (x+3)2=−1{\left( {x + 3} \right)^2} = - 1, we get,
(x+3)=−1\left( {x + 3} \right) = \sqrt { - 1}
Therefore, using (1)\left( 1 \right),
⇒(x+3)=±i\Rightarrow \left( {x + 3} \right) = \pm i
Subtracting 3 from both the sides,

⇒x=±i−3 \Rightarrow x = \pm i - 3

Thus, this is the required answer.

Note: If in the question, it was not mentioned that we have to use the method of completing the square, then, we could have used the quadratic formula to solve the given quadratic equation.
Given quadratic equation is x2+6x+10=0{x^2} + 6x + 10 = 0
Comparing this with the general quadratic equation i.e. ax2+bx+c=0a{x^2} + bx + c = 0
We have,
a=1a = 1, b=6b = 6 and c=10c = 10
Now, we can find the roots of a quadratic equation using the quadratic formula, x=−b±b2−4ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
Hence, for the equation x2+6x+10=0{x^2} + 6x + 10 = 0, substituting a=1a = 1, b=6b = 6 and c=10c = 10, we get,
x=−6±62−4(1)(10)2(1)x = \dfrac{{ - 6 \pm \sqrt {{6^2} - 4\left( 1 \right)\left( {10} \right)} }}{{2\left( 1 \right)}}
⇒x=−6±36−402=−6±−42\Rightarrow x = \dfrac{{ - 6 \pm \sqrt {36 - 40} }}{2} = \dfrac{{ - 6 \pm \sqrt { - 4} }}{2}
Solving further, we get,
⇒x=−6±2−12=−2(3±−1)2\Rightarrow x = \dfrac{{ - 6 \pm 2\sqrt { - 1} }}{2} = \dfrac{{ - 2\left( {3 \pm \sqrt { - 1} } \right)}}{2}
Here, substituting −1=±i\sqrt { - 1} = \pm i as it is imaginary, and cancelling out the same terms from the numerator and denominator, we get,
⇒x=−3±i\Rightarrow x = - 3 \pm i
Thus, this is the required answer.