Question

If one of the roots of the equation ${{x}^{2}}+ax-b=0$ is 1, then what is ( a – b) equal to?
( a ) – 1
( b ) 1
( c ) 2
( d ) - 2

Answer 1

Hint: In this question, we are given a quadratic equation with one of its roots and we have to find the value of (a – b). First, we compare the equation with the standard form of a quadratic equation and find out the value of a, b and c then we suppose the roots of the quadratic equation and after comparing the roots with the sum of roots and the product of roots, we find the value of ( a – b)

Formula Used: Sum of roots = $\dfrac{-b}{a}$
 Product of roots = $\dfrac{c}{a}$

Complete step by step Solution:
Given quadratic equation is ${{x}^{2}}+ax-b=0$………………………. (1)
To find the roots of a quadratic equation we compare the equation (1) with the standard quadratic equation $a{{x}^{2}}+bx+c=0$, we get
a = 1 , b = a and c = -b
Let p and q be the roots of the given quadratic equation.
Then sum of roots ( p + q) = $-\dfrac{b}{a}=-\dfrac{a}{1}$ = - a
And the product of roots (pq) = $\dfrac{c}{a}=-\dfrac{b}{1}$= -b
We are given one of the roots of equation 1.
Let q = 1 then
p+q = -a
And p + 1 = -a
That is a = -p -1
Similarly pq = -b
That is p(1) = -b
Then b = -p
We have to find the value of a – b
That is a – b = ( -p-1) – (-p)
     a – b = -p-1+p = - 1
Hence, the value of ( a – b ) = - 1

Therefore, the correct option is (a).

Note: Whenever we solve these types of questions where roots are given, we use the identity of the product of roots which is if x and y are the roots of any quadratic equation the value of xy will be equal to $\dfrac{ constant\, term}{coefficient\, of\, x^2}$ and sum of the roots that is x + y is equal to $\dfrac{ -coefficient\, of \,x}{coefficient\, of\, x^2}$ and by solving it we get the desired answer.