Question

How do you solve \[7{x^2} - 5 = 2x + 9{x^2}\] using the quadratic formula?

Answer 1

Hint: We first need to rearrange the given equation in the standard form of quadratic equation. That is in the form of \[a{x^2} + bx + c = 0\]. After that we can solve this using various methods that are by completing the square, factorization, graph or by quadratic formula. Here we need to use a quadratic formula that is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

Complete step-by-step solution:
Given, \[7{x^2} - 5 = 2x + 9{x^2}\]
Shifting the terms we have,
\[2x + 9{x^2} - 7{x^2} + 5 = 0\]
\[2{x^2} - 2x + 5 = 0\]
Since the degree of the equation is 2, we have 2 factors or two roots.
On comparing the given equation with the standard quadratic equation\[a{x^2} + bx + c = 0\], we have\[a = 2\], \[b = - 2\] and \[c = 5\].
Now we use quadratic formula or Sridhar’s formula,
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Substituting we have,
\[ \Rightarrow x = \dfrac{{ - ( - 2) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4(2)(5)} }}{{2(2)}}\]
\[ = \dfrac{{2 \pm \sqrt {4 - 40} }}{4}\]
\[ = \dfrac{{2 \pm \sqrt { - 36} }}{4}\]
\[ = \dfrac{{2 \pm \sqrt { - 1 \times 36} }}{4}\]
We know that \[\sqrt { - 1} = i\],
\[ = \dfrac{{2 \pm i\sqrt {36} }}{4}\]
We know that 36 is a perfect square,
\[ = \dfrac{{2 \pm 6i}}{4}\]
Taking 2 common,
\[ = \dfrac{{2(1 \pm 3i)}}{4}\]
\[ = \dfrac{{1 \pm 3i}}{2}\]
Thus we have two roots,
\[ \Rightarrow x = \dfrac{{1 + 3i}}{2}\] and \[x = \dfrac{{1 - 3i}}{2}\]. This is the required answer.

Note: Since we have a polynomial of degree two and hence it is called quadratic polynomial. If we have a polynomial of degree ‘n’ then we have ‘n’ roots. In the given problem we have a degree that is equal to 2. Hence the number of roots are 2. Also we know that \[\sqrt { - 1} \] is undefined and we take \[\sqrt { - 1} = i\] that is an imaginary number. Quadratic formula and Sridhar’s formula are both the same. We know that the product of two negative numbers gives us a positive number. Also keep in mind when shifting values from one side of the equation to another side of the equation, always change sign from positive to negative and vice-versa.