Question

The roots of the equation \[\left( {x - a} \right)\left( {x - b} \right) = {b^2}\] are
A.Real and equal
B.Real and unequal
C.Imaginary
D.Equal

Answer 1

Hint: We are given with an equation and need to find the nature of its roots. We know that nature of the roots is determined from its discriminant of the form \[D = {b^2} - 4ac\] . the nature of the roots are
1.If \[{b^2} - 4ac < 0\] then the equation has no real root.
2.If \[{b^2} - 4ac = 0\] then the equation has equal and real roots.
3.If \[{b^2} - 4ac > 0\] then the equation has unequal but real roots.
From these conditions we can determine the nature of the roots of the above equation. But very first we have to write it in the form \[a{x^2} + bx + c = 0\].

Complete step-by-step answer:
Given that,
\[\left( {x - a} \right)\left( {x - b} \right) = {b^2}\]
On multiplying the brackets we get,
\[{x^2} - bx - ax + ab = {b^2}\]
Taking x common from middle two terms,
\[{x^2} - x\left( {a + b} \right) + ab = {b^2}\]
\[{x^2} - x\left( {a + b} \right) + ab - {b^2} = 0\]
This is of the form \[a{x^2} + bx + c = 0\].
Now let’s find the discriminant D.
\[D = {b^2} - 4ac\]
Putting the values of a,b and c
\[ \Rightarrow {\left( {a + b} \right)^2} - 4 \times 1 \times \left( {ab - {b^2}} \right)\]
Expanding the first bracket and multiplying the terms in second bracket
\[ \Rightarrow {a^2} + 2ab + {b^2} - 4ab + 4{b^2}\]
Rearranging the terms we get,
\[ \Rightarrow {a^2} - 2ab + {b^2} + 4{b^2}\]
Now if we observe that first three terms is expansion of an algebraic identity,
\[ \Rightarrow {\left( {a - b} \right)^2} + {\left( {2b} \right)^2}\]
This is our discriminant D.
\[D \Rightarrow {\left( {a - b} \right)^2} + {\left( {2b} \right)^2}\]
Since it is always greater than zero this equation has real but distinct roots.
Thus option B is the correct option.

Note: Students might get confused how we should decide the value of D without getting any value of a and b. So here we have intentionally written the terms in square form. Because we know that a square term is always positive and D is greater than zero because two terms are added here.