Question

Solve the given equation $ {x^2} + x - (a + 2)(a + 1) = 0 $ by using a quadratic equation formula.

Answer 1

Hint: The Quadratic Formula for quadratic equation $a`{x^2} + bx + c = 0$ is given by
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4a`c} }}{{2a`}}$. Here we will use $a'$ in place of ‘a’ because ‘a’ is already used in question.

Complete step-by-step solution:
The given quadratic equation is ${x^2} + x - (a + 2)(a + 1) = 0$
On comparing the above equation with standard quadratic equation $a`{x^2} + bx + c = 0$,
we get
$
  a` = 1 \\
  b = 1 \\
  c = - (a + 2)(a + 1) \\
 $
On putting values of $a`,b,c$ in Quadratic Formula we get
 $
  x = \dfrac{{ - b \pm \sqrt {{b^2} - 4a`c} }}{{2a`}} \\
   \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 + 4(a + 2)(a + 1)} }}{2} \\
    \\
$ _____________________________ eq 1.

$ {\text{let }}{D^2} = 1 + 4(a + 2)(a + 1) $
On solving above equation, we get
\[\
   \Rightarrow {D^2} = 1 + 4{a^2} + 12a + 8 \\
   \Rightarrow {D^2} = 4{a^2} + 12a + 9 \\
   \Rightarrow {D^2} = (2a + 3)(2a + 3) \\
   \Rightarrow {D^2} = {(2a + 3)^2} \\
    \\
 \]
Now put the value of ${D^2}$ in eq 1. We get
\[
   \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {{D^2}} }}{2} \\
   \Rightarrow x = \dfrac{{ - 1 \pm (2a + 3)}}{2} \\
 \]
On Solving above equation separately for + sign we get
\[\
   \Rightarrow x = \dfrac{{ - 1 + 2a + 3}}{2} \\
   \Rightarrow x = \dfrac{{2a + 2}}{2} \\
   \Rightarrow x = a + 1 \\
    \\
 \]
Now for solving – sign , we get
$
   \Rightarrow x = \dfrac{{ - 1 - (2a + 3)}}{2} \\
   \Rightarrow x = \dfrac{{ - 2a - 4}}{2} \\
   \Rightarrow x = - a - 2 \\
 $
Hence roots of given quadratic equation are
$x = a + 1{\text{ or }}x = - (a + 2)$

Note: Whenever you get this type of question the key concept of solving is you have to just compare this equation with the general form of the quadratic equation then put the values of a,b,c according to the quadratic equation in quadratic formula and then solve it to get the answer.