
The value of n for which the expression $9{x^4} - 12{x^3} + n{x^2} - 8x + 4$ becomes a perfect square is
(A) 12
(B) 16
(C) 18
(D) 24
Answer
487.2k+ views
Hint:To solve this type of particular problem we use algebraic expression of square term.Let assume $9{x^4} - 12{x^3} + n{x^2} - 8x + 4$ = ${\left( {a{x^2} + bx + c} \right)^2}$.Expand the R.H.S term i.e assumed value by using formula ${\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2xy + 2yz + 2zx$.After substituting assumed value in formula we have to compare the similar terms to get our answer.
Complete step-by-step answer:
Let assume $9{x^4} - 12{x^3} + n{x^2} - 8x + 4$ = ${\left( {a{x^2} + bx + c} \right)^2}$
We assume this because the question says $9{x^4} - 12{x^3} + n{x^2} - 8x + 4$ is a perfect square.
Now we expand ${\left( {a{x^2} + bx + c} \right)^2}$by using formula ${\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2xy + 2yz + 2zx$
So ${\left( {a{x^2} + bx + c} \right)^2} = {a^2}{x^4} + {b^2}{x^2} + {c^2} + 2ab{x^3} + 2bcx + 2ac{x^2}$
Now ${a^2}{x^4} + {b^2}{x^2} + {c^2} + 2ab{x^3} + 2bcx + 2ac{x^2}$=$9{x^4} - 12{x^3} + n{x^2} - 8x + 4$
So now we compare terms
Firstly comparing ${x^4}$ we get ${a^2} = 9$
And $a = \pm 3$
Now compare constant term ${c^2} = 4$
And $c = \pm 2$
Now comparing ${x^3}$ term we get
$2ab = - 12$
And $ab = - 6$
Now we have to put value of $a$ so that we get
$b = \pm 2$
Now comparing ${x^2}$ we get
$2ac + {b^2} = n$
Now we have to put the value of a,b,c in the above equation to find the value of n.
$2 \times 3 \times 2 + {\left( { \pm 2} \right)^2}=n$
So the square of any number is always positive
$12 + 4$
$n = 16$
The value of $n = 16$
So, the correct answer is “Option B”.
Note:In this type of question we have to remember all the algebraic expressions and formulas that how we can use them .We have to compare all the similar terms and remember whenever we solve a square term we have to always use \[ \pm \] both signs so that we get all possible values that we need.
Complete step-by-step answer:
Let assume $9{x^4} - 12{x^3} + n{x^2} - 8x + 4$ = ${\left( {a{x^2} + bx + c} \right)^2}$
We assume this because the question says $9{x^4} - 12{x^3} + n{x^2} - 8x + 4$ is a perfect square.
Now we expand ${\left( {a{x^2} + bx + c} \right)^2}$by using formula ${\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2xy + 2yz + 2zx$
So ${\left( {a{x^2} + bx + c} \right)^2} = {a^2}{x^4} + {b^2}{x^2} + {c^2} + 2ab{x^3} + 2bcx + 2ac{x^2}$
Now ${a^2}{x^4} + {b^2}{x^2} + {c^2} + 2ab{x^3} + 2bcx + 2ac{x^2}$=$9{x^4} - 12{x^3} + n{x^2} - 8x + 4$
So now we compare terms
Firstly comparing ${x^4}$ we get ${a^2} = 9$
And $a = \pm 3$
Now compare constant term ${c^2} = 4$
And $c = \pm 2$
Now comparing ${x^3}$ term we get
$2ab = - 12$
And $ab = - 6$
Now we have to put value of $a$ so that we get
$b = \pm 2$
Now comparing ${x^2}$ we get
$2ac + {b^2} = n$
Now we have to put the value of a,b,c in the above equation to find the value of n.
$2 \times 3 \times 2 + {\left( { \pm 2} \right)^2}=n$
So the square of any number is always positive
$12 + 4$
$n = 16$
The value of $n = 16$
So, the correct answer is “Option B”.
Note:In this type of question we have to remember all the algebraic expressions and formulas that how we can use them .We have to compare all the similar terms and remember whenever we solve a square term we have to always use \[ \pm \] both signs so that we get all possible values that we need.
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