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Q.

Solve the following equation for x.


1 a+b+x = 1 a + 1 b + 1 x ;a≠b≠0,x≠0,x≠−(a+b)


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a

The value of x is −b,a

b

The value of x is b,−a

c

The value of x is b,a

d

The value of x is −b,−a  

answer is D.

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Detailed Solution

Given, the quadratic equation 1 a+b+x = 1 a + 1 b + 1 x ;a≠b≠0,x≠0,x≠−(a+b). We know that for a quadratic equation,
(i) If the discriminant is positive, two separate roots exist.
(ii) Both roots are equivalent if the discriminant is zero.
(iii) There are no real roots if the discriminant is negative. Instead, there are two different (non-real) complex roots.
In quadratic equation, a x 2 +bx+c=0(a≠0) Discriminant, D= b 2 −4ac .
Simplify the given equation.
Question ImageLet us find the Discriminant.
Question Image Now, use quadratic formula to find the roots.
  x= − k 2 ± k 2 (a−b) 2 2(k) x= − k 2 ±k(a−b) 2k x= k 2k (−k±(a−b)) x= 1(−(a+b)±(a−b)) 2 x= (−a−b+a−b) 2  or  (−a−b−a+b) 2 x= −2b 2  or  −2a 2 x=−b or −a The value of x is −b,−a.
The correct option is (4).
 
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