Question

How do you solve $ - 4{x^2} + x + 9 = 0 $ using the quadratic formula?

Answer 1

Hint: The given equation is a quadratic equation in one variable $ x $ . The general form of a quadratic equation is given by $ a{x^2} + bx + c = 0 $ . Solving this equation gives two values of the variable $ x $ as the result. We can solve this equation by using the quadratic formula $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ .

Complete step by step solution:
We have to solve the given equation $ - 4{x^2} + x + 9 = 0 $ using the quadratic formula $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
To find the value of $ x $ , we have to put the values of $ a $ , $ b $ and $ c $ in the quadratic formula. To get the values of $ a $ , $ b $ and $ c $ from the given equation, we rearrange the equation and compare it with the general form of the quadratic equation.
General form of quadratic equation is written in the form of $ a{x^2} + bx + c = 0 $ , where $ a $ is the coefficient of $ {x^2} $ , $ b $ is the coefficient of $ x $ and $ c $ is the constant term. The RHS is $ 0 $ .
On comparing the above rearranged equation with the general form, we observe that
Co-efficient $ a $ of $ {x^2} $ is $ - 4 $ ,
Co-efficient $ b $ of $ x $ is $ 1 $ ,
and the constant term $ c $ is $ 9 $ .
Thus, $ a = - 4 $ , $ b = 1 $ and $ c = 9 $ .
Now we put the values of $ a $ , $ b $ and $ c $ in the quadratic formula to solve for value of $ x $ .
\[
  x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
   \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {{{(1)}^2} - 4 \times ( - 4) \times 9} }}{{2 \times ( - 4)}} \\
   \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 - ( - 144)} }}{{ - 8}} \\
   \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 + 144} }}{{ - 8}} \\
   \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {145} }}{{ - 8}} \\
\]
Multiplying $ - 1 $ in the numerator and the denominator, we get:
\[
   \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {145} }}{{ - 8}} \times \dfrac{{ - 1}}{{ - 1}} \\
   \Rightarrow x = \dfrac{{1 \pm \sqrt {145} }}{8} \\
\]
In simplified form,
\[x = \dfrac{{1 + \sqrt {145} }}{8}or\dfrac{{1 - \sqrt {145} }}{8}\]
\[x = \dfrac{{1 + \sqrt {145} }}{8}\]
Thus, the two values of $ x $ that we get on solving the given equation are \[\dfrac{{1 + \sqrt {145} }}{8}\] and \[\dfrac{{1 - \sqrt {145} }}{8}\].

Note: Another method to solve for $ x $ in the quadratic equation is by factorization. Using quadratic formula is simpler than factorization as it involves direct calculation using values of $ a $ , $ b $ and $ c $ . We get two values of $ x $ while solving the quadratic equation. We can check the answer by putting the result in the given equation to satisfy LHS = RHS.