Question

If the zeroes of the polynomial \[f\left( x \right)={{x}^{3}}-12{{x}^{2}}+39x+k\] are in AP, then the value of k is:
(a) 28
(b) – 28
(c) 30
(d) – 30

Answer 1

Hint: Assume the zeroes of the polynomial f(x) as a, a + d, a + 2d where ‘a’ is the first term of the AP and ‘d’ is its common difference. Now, assume the coefficient of \[{{x}^{3}}\] as A, \[{{x}^{2}}\] as B and x as C and the constant term as D in the polynomial f(x). Apply the formulas: \[\text{Sum of roots}=\dfrac{-B}{A}\] and the sum of the product of roots taken two at a time \[=\dfrac{C}{A}\] to firm two relations between ‘a’ and ‘d’. Then solve the equation to find ‘a’ and ‘d’. Finally, substitute the obtained values in the formula: \[\text{product of roots}=\dfrac{-D}{A}\] to find the value of k.

Complete step-by-step solution:
Here, we have been provided with the information that the roots of the polynomial \[f\left( x \right)={{x}^{3}}-12{{x}^{2}}+39x+k\] are in AP and we have to find the value of k. Now, let us assume that the roots or zeros of the polynomial f(x) are a, a + d, a + 2d, where ‘a’ is the first term of the AP and ‘d’ is its common difference. Here, we have assumed three roots because the f(x) is a cubic polynomial, so the maximum number of real roots will be three.
We know that for a cubic polynomial, \[A{{x}^{3}}+B{{x}^{2}}+Cx+D\] whose roots are \[\alpha ,\beta ,\gamma \] we have,
\[\text{Sum of roots}=\alpha +\beta +\gamma =\dfrac{-B}{A}\]
Sum of the product of roots taken two at a time \[=\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{C}{A}\]
\[\text{Product of roots}=\alpha \beta \gamma =\dfrac{D}{A}\]
So, comparing \[f\left( x \right)={{x}^{3}}-12{{x}^{2}}+39x+k\] with \[A{{x}^{3}}+B{{x}^{2}}+Cx+D\] we get,
A = 1
B = – 12
C = 39
D = k
Now, comparing the roots, we get,
\[\alpha =a\]
\[\beta =a+d\]
\[\gamma =a+2d\]
Therefore, applying the identities listed above, we have,
\[\Rightarrow \alpha +\beta +\gamma =\dfrac{-B}{A}\]
\[\Rightarrow a+a+d+a+2d=\dfrac{-\left( -12 \right)}{1}\]
\[\Rightarrow 3a+3d=12\]
\[\Rightarrow a+d=4.......\left( i \right)\]
Now, we have,
\[\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{C}{A}\]
\[\Rightarrow a\left( a+d \right)+\left( a+d \right)\left( a+2d \right)+\left( a+2d \right)a=\dfrac{39}{1}\]
\[\Rightarrow 3{{a}^{2}}+6ad+2{{d}^{2}}=39......\left( ii \right)\]
Now, substituting the value of ‘d’ from equation (i) in equation (ii), we get,
\[\Rightarrow 3{{a}^{2}}+6a\left( 4-a \right)+2{{\left( 4-a \right)}^{2}}=39\]
\[\Rightarrow 3{{a}^{2}}+24a-6{{a}^{2}}+2\left( 16+{{a}^{2}}-8a \right)=39\]
\[\Rightarrow 3{{a}^{2}}+24a-6{{a}^{2}}+32+2{{a}^{2}}-16a=39\]
\[\Rightarrow -{{a}^{2}}+8a-7=0\]
\[\Rightarrow {{a}^{2}}-8a+7=0\]
On using the middle term split method, we get,
\[\Rightarrow {{a}^{2}}-7a-a+7=0\]
\[\Rightarrow \left( a-7 \right)\left( a-1 \right)=0\]
\[\Rightarrow a=7;a=1\]
Considering each value of ‘a’ one by one, we have,
Case (1): When a = 7,
\[\Rightarrow a+d=4\]
\[\Rightarrow 7+d=4\]
\[\Rightarrow d=4-7\]
\[\Rightarrow d=-3\]
Now, applying the formula for the product of roots given as \[\alpha \beta \gamma =\dfrac{-D}{A},\] we get,
\[\Rightarrow a\times \left( a+d \right)\times \left( a+2d \right)=\dfrac{-k}{1}\]
\[\Rightarrow 7\times 4\times \left( 7+2\times \left( -3 \right) \right)=-k\]
\[\Rightarrow 28\times 1=-k\]
\[\Rightarrow k=-28\]
Case (2): When a = 1
\[\Rightarrow a+d=4\]
\[\Rightarrow 1+d=4\]
\[\Rightarrow d=4-1\]
\[\Rightarrow d=3\]
Again applying the formula for the product of roots given as \[\alpha \beta \gamma =\dfrac{-D}{A},\] we get,
\[\Rightarrow a\times \left( a+d \right)\times \left( a+2d \right)=\dfrac{-k}{1}\]
\[\Rightarrow 1\times 4\times \left( 1+2\times 3 \right)=-k\]
\[\Rightarrow 1\times 4\times 7=-k\]
\[\Rightarrow 28=-k\]
\[\Rightarrow k=-28\]
Hence, option (b) is the right answer.

Note: One may note that in both cases we obtained the same value of k. This is proof that our answer is correct. In the above question, you must remember the formulas of the sum and product of roots to solve it. Never assume the three roots of the polynomial as three different variables in the above question otherwise you will get confused in the calculation. Note that to reduce the calculation you may assume the three roots in AP as a – d, a, a + d.