Question

Find the solution of \[9{x^2} + 6x + 1 = 0\] using quadratic formula.

Answer 1

Hint: The quadratic formula is derived from the process of completing the square, and is formally stated as, for \[a{x^2} + bx + c = 0\] , the values of x which are the solutions of the equation are given by \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] . For the quadratic formula to work the equation must be arranged in the form quadratic=0.

Complete answer:
The quadratic formula states that for \[a{x^2} + bx + c = 0\] , the values of x which are the solutions to the equation are given by:
 \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Since the quadratic equation which we have is \[9{x^2} + 6x + 1 = 0\] . This means that the values of a, b and c are 9, 6 and 1 respectively.
Now, substituting 9 for a; 6 for b and 1 for c in the above equation gives,
 \[x = \dfrac{{ - 6 \pm \sqrt {{6^2} - \left( {4 \times 9 \times 1} \right)} }}{{2.9}}\]
The square root or the symbol of plus minus should not be dropped in between. \[{b^2}\] Means the square of all b, including its sign, so \[{b^2}\] should not be left even if b is negative because the square of a negative is positive.
Now solving we have,
 \[
   \Rightarrow x = \dfrac{{ - 6 \pm \sqrt {36 - 36} }}{{2 \times 9}} \\
   \Rightarrow x = \dfrac{{ - 6 \pm \sqrt 0 }}{{2 \times 9}} \\
   \Rightarrow x = \dfrac{{ - 6 \pm 0}}{{2 \times 9}} \\
   \Rightarrow x = \dfrac{{ - 6}}{{18}} \\
\]
Converting the value of x in lowest form we have,
 \[ \Rightarrow x = - \dfrac{1}{3}\]
Hence, the quadratic formula \[9{x^2} + 6x + 1 = 0\] is solved and the value of x is \[ - \dfrac{1}{3}\] .

Note: Often, the simplest way to solve \[a{x^2} + bx + c = 0\] for the value of x is to factor the quadratic, set each factor equal to zero, and then solve each factor. But sometimes the quadratic is too messy, or it does not factor at all. While factoring may not always be successful, the quadratic formula can always find the solution.