Question

The number of values of the pair (a, b) for which $ a{(x + 1)^2} + b({x^2} - 3x - 2) + x + 1 = 0,\forall x \in R $ .
A. 0
B. 1
C. 2
D. infinite

Answer 1

Hint: The expression in the above question is a quadratic expression and we know that the roots of a quadratic expression can be calculated by methods like completing the squares or finding the discriminant and using the discriminant to find the value of the roots.

Complete step-by-step answer:
Since we have to calculate the values of the pair (a,b) in the following expression. We need to expand the value of (x+1) whole square to get
 $ a({x^2} + 1 + 2x) + b({x^2} - 3x - 2) + x + 1 = 0 $
Rearranging the terms so as to get all the terms of $ {x^2} $ together and have a common coefficient to $ {x^2} $ by expanding all the terms,
 $
  a{x^2} + a + 2ax + b{x^2} - 3bx - 2b + x + 1 = 0 \\
  {x^2}(a + b) + x(2a - 3b + 1) + a - 2b + 1 = 0 \;
 $
Now we have arranged all the terms properly as we observe in a general quadratic expression and we know that the value of the above quadratic expression has been given as 0 and all the x belongs to real numbers. Therefore if the value of x will be 0 then value of a and b will give us three equations as,
 $
  2a - 3b + 1 = 0 \\
  a + b = 0 \\
  a - 2b + 1 = 0 \;
 $
Therefore from the last equation we have the value of a as
 $
  a - 2b + 1 = 0 \\
  a = 2b - 1 \;
 $
From above the value of b can be calculated from a as
 $
\Rightarrow a + b = 0 \\
\Rightarrow b = - a \\
\Rightarrow b = - (2b - 1) \\
\Rightarrow b = 1 - 2b \\
\Rightarrow 3b = 1 \\
\Rightarrow b = \dfrac{1}{3} \\
 $
Similarly a will be by putting the value of b
 $ \Rightarrow a = 2b - 1 = 2\left( {\dfrac{1}{3}} \right) - 1 = \dfrac{2}{3} - 1 = \dfrac{{2 - 3}}{3} = \dfrac{{ - 1}}{3} $
On putting these values of a and b in the equation 2a-3b+1=0 we get,
 $
\Rightarrow 2\left( { - \dfrac{1}{3}} \right) + 3\left( {\dfrac{1}{3}} \right) - 1 = 0 \\
\Rightarrow - \dfrac{2}{3} + 1 - 1 = 0 \\
\Rightarrow - \dfrac{2}{3} = 0 \\
 $
But we know that $ - \dfrac{2}{3} \ne 0 $ hence there is no value of (a,b) for the given quadratic expression when x belongs to real numbers.
So, the correct answer is “0”.

Note: Hence we can conclude from the above that (a,b)=φ that is it is empty and has zero values for the above quadratic expression when all the x belongs to real numbers.