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Let `a,b in N, a != b` and the two quadratic equations `(a-1)x^2-(a^2+2)x+a^2+2a=0 and (b-1)x^2-(b^2+2)x+(b^2+2b)=0` have a common root. The value of `ab` isA. `4`B. `6`C. `8`D. `10 |
Answer» Correct Answer - C `(c )` `a gt 1`, `b gt 1`, `a ne b` Let `alpha `is the common root `(a-1)alpha^(2)-(a^(2)+1)alpha+a^(2)+2a=0` ………`(i)` `(b-1)alpha^(2)-(b^(2)+2)alpha+(b^(2)+2b)=0` ………..`(ii)` `(i)xx(b-1)-(ii)xx(a-1)`, we get `(a-b)(ab-a-b-2)(alpha-1)=0`..........`(iii)` If `alpha=1`, then `a=b=1` So, `a ne 1` From `(iii)`, we get `ab=a+b+2` Let `a gt b gt 1` `:. b=1+(b)/(a)+(2)/(a) lt 3 implies b=2`, `a=4` `:. ab=8` |
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