InterviewSolution
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InterviewSolution
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Let `f(x)=3x^(2)-5x-1`. Then solve f(x)=0 by (i) factroing the quadratic (ii) using th quadratic formula (iii) completing the square and then rewrite f(x) in the form `A(x+-B)^(2)+-C`. |
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Answer» `f(x)=3x^(2)-5x-1` :.f(x)=0 `implies3x^(2)-5x-1=0` (i) The given quadratic equation cannot be fully factorised using real integers. So it is better to solve this equation by any other method. (ii) `3x^(2)-5x-1=0` Compare it with `ax^(2)+bx+c=0`, we get a=3, b=-5, c=-1 :. Let two roots of this equation are `alpha=(-b+sqrt(b^(2)-4ac))/(2a)andbeta=(-b-sqrt(b^(2)-4ac))/(2a)` `(-(5)+sqrt((-5)^(2)-4(3)(-1)))/(2xx3)` `=(5+sqrt37)/(6)` `=(-(-5)-sqrt((-5)^(2)-4(3)(-1)))/(2xx3)` `beta=(5-sqrt(37))/(6)` :. Two values of x are `(5+sqrt(37))/(6)and(5-sqrt37)/(6)` (iii) `3x^(2)-5x-1=0` `impliesx^(2)-(5)/(3)x(1)/(3)=0" "("dividing both sides by 3")` `impliesx^(2)-(5)/(3)x+.....=(1)/(3)+.....` `impliesx^(2)-(5)/(3)x+((5)/(6))^(2)=(1)/(3)+((5)/(6))^(2)" "["adding"(("coeff. of x")^(2)/(2))"on both sides"]` `implies(x-(5)/(6))^(2)=(1)/(3)+(25)/(36)` `implies(x-(5)/(6))^(2)=(12+25)/(36)` `implies(x-(5)/(6))^(2)=((sqrt37)/(6))^(2)` `impliesx-(5)/(6)=+-(sqrt37)/(6)` `:.x=(5)/(6)+(sqrt37)/(6)=(5+sqrt37)/(6),x=(5)/(6)-(sqrt37)/(6)=(5-sqrt37)/(6)` :. Two values of x are `(5+sqrt37)/(6)and(5-sqrt(37))/(6)`. Now, `f(x)=3x^(2)-5x-1=3(x^(2)-(5)/(3)x)-1` `=3(x^(2)-(5)/(3)x+(25)/(36)-(25)/(36))-1` `=3(x-(5)/(6))^(2)-(25)/(12)-1=3(x-(5)/(6))^(2)-(37)/(12)` which is of the form `A(x-B)^(2)-C`, where A=3, `B=(5)/(6),C=(37)/(12)`. |
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