Question

How do you solve \[{x^2} + 2x - 1 = 2\] using the quadratic formula?

Answer 1

Hint: A polynomial of degree two is called a quadratic polynomial and its zeros can be found using many methods like factorization, completing the square, graphs, quadratic formula etc. The quadratic formula is used when we fail to find the factors of the equation. In the given question we have to solve the given quadratic equation using the quadratic formula. That is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

Complete step-by-step solution:
Given, \[{x^2} + 2x - 1 = 2\].
The given equation should be in standard form \[a{x^2} + bx + c = 0\].
\[
  {x^2} + 2x - 1 - 2 = 0 \\
  {x^2} + 2x - 3 = 0 \\
\]
On comparing the given equation with the standard quadratic equation \[a{x^2} + bx + c = 0\]. We get, \[a = 1\], \[b = 2\] and \[c = - 3\].
Substituting in the formula of standard quadratic, \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
\[ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4(1)( - 3)} }}{{2(1)}}\]
We know that the product of two negative numbers gives a positive number.
\[ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt {4 + 12} }}{2}\]
\[ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt {16} }}{2}\]
We know that the square root of 16 is 4.
\[ \Rightarrow x = \dfrac{{ - 2 \pm 4}}{2}\]
Thus we have two roots
\[ \Rightarrow x = \dfrac{{ - 2 + 4}}{2}\] and \[x = \dfrac{{ - 2 - 4}}{2}\]
\[ \Rightarrow x = \dfrac{2}{2}\] and \[x = \dfrac{{ - 6}}{2}\]
\[ \Rightarrow x = 1\] and \[x = - 3\]

Therefore the roots are x =1 and x = 3.

Additional information: In various fields of mathematics require the point at which the value of a polynomial is zero, those values are called the factors/solution/zeros of the given polynomial. On the x-axis, the value of y is zero so the roots of an equation are the points on the x-axis, that is the roots are simply the x-intercepts.

Note: We can also solve this using simple factorization. That is the middle term ‘2x’ can be expanded as ‘-3x’ and ‘x’ (Because the multiplication of 3 and -1 is -3. If we multiply 3 and -1 we get -3 and if we add 3 and -1 we get 2). That is \[ \Rightarrow {x^2} + 2x - 3 = 0\]
\[ \Rightarrow {x^2} + 3x - x - 3 = 0\]
Taking ‘x’ common on the first two terms and -1 common on the last two terms we have,
\[ \Rightarrow x(x + 3) - 1(x + 3) = 0\]
Taking \[(x + 3)\] common on the above equation we have,
\[ \Rightarrow (x + 3)(x - 1) = 0\].
Using zero product principle we have,
\[ \Rightarrow (x + 3) = 0\] and \[(x - 1) = 0\]
\[ \Rightarrow x = - 3\] and \[x = 1\]. In both the cases we have the same answer.