If a < b < c < d, then the roots of fthe equation (x – a) (x – c) + 2(x – b) (x – d) = 0 are [ ] a) non real complex b) real and equal c) real and unequal d) none of these
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We have,
(x−a)(x−c)+2(x−b)(x−d)=0
⇒3x2−(a+c+2b+2d)x+(ac+2bd)=0
Discriminant
={(a+2d)+(c+2b)}2−4.3(ac+2bd)
={(a+2d)+(c+2b)}2−12(ac+2bd)
={(a+2d)−(c+2b)}2−4(a+2d)(c+2b)−12(ac+2bd)
={(a+2d)−(c+2b)}2+4ac+8ab+8cd+16bd−12ac−24bd
={(a+2d)−(c+2b)}2+8(ab+cd−ac−bd)
={(a+2d)−(c+2b)}2+8(c−b)(d−a)>0
∵a<b<c<d⇒c−b>
Hence the roots of this equation are real and unequal
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