Solve \[{x^4} + {x^3} - 16{x^2} - 4x + 48 = 0\] given that the product of its two roots is 6.
Answer
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Hint:
Here, we need to find the roots of the given quartic equation. We will find the product of the other roots of the quartic equation. Then, using Descartes’s solution, we will rewrite the quartic equation as a product of two quadratic equations. Finally, we will factorise the two quadratic equations to get the four roots of the given quartic equation.
Formula Used:
We will use the following formulas:
1) The product of all roots of a quartic equation of the form \[a{x^4} + b{x^3} + c{x^2} + dx + e = 0\] is given by \[\dfrac{e}{a}\].
2) According to Descartes’s solution, we can write a quartic equation \[a{x^4} + b{x^3} + c{x^2} + dx + e = 0\] as \[{x^4} + b{x^3} + c{x^2} + dx + e = \left( {{x^2} + sx + t} \right)\left( {{x^2} + ux + v} \right)\], which can be further simplified as \[\left( {{x^2} + sx + t} \right)\left( {{x^2} + ux + v} \right) = {x^4} + \left( {s + u} \right){x^3} + \left( {t + v + su} \right){x^2} + \left( {sv + tu} \right)x + tv\]. Here, \[b = s + u\], \[c = t + v + su\], \[d = sv + tu\], and \[e = tv\].
Complete step by step solution:
Let the four roots of the given quartic equation be \[\alpha \], \[\beta \], \[\gamma \], and \[\delta \].
It is given that the product of two roots is 6.
Thus, we can write
\[\alpha \beta = 6\]
Now, the product of all roots of a quartic equation of the form \[a{x^4} + b{x^3} + c{x^2} + dx + e = 0\] is given by \[\dfrac{e}{a}\].
Comparing the equation \[{x^4} + {x^3} - 16{x^2} - 4x + 48 = 0\] to the equation \[a{x^4} + b{x^3} + c{x^2} + dx + e = 0\], we get
\[a = 1\] and \[e = 48\]
Thus, we get the product of all roots of the quartic equation as
\[\dfrac{e}{a} = \dfrac{{48}}{1} = 48\]
The roots are \[\alpha \], \[\beta \], \[\gamma \], and \[\delta \].
Therefore, we can write
\[\alpha \beta \gamma \delta = 48\]
Substituting \[\alpha \beta = 6\] in the equation, we get
\[ \Rightarrow 6\gamma \delta = 48\]
Dividing both sides by 6, we get
\[ \Rightarrow \gamma \delta = 8\]
Using Descartes’s solution, we can factorise the quartic equation into two quadratic equations.
Comparing the coefficients of the terms in the equations \[{x^4} + {x^3} - 16{x^2} - 4x + 48 = 0\] and \[{x^4} + \left( {s + u} \right){x^3} + \left( {t + v + su} \right){x^2} + \left( {sv + tu} \right)x + tv\], we get
\[s + u = 1\]
\[sv + tu = - 4\]
\[tv = 48\]
The constant 48 is the product of 8 and 6.
Therefore, substituting \[t = 8\] and \[v = 6\], we get
\[6s + 8u = - 4\]
Multiplying both sides of the equation \[s + u = 1\] by 6, we get
\[\begin{array}{l} \Rightarrow 6\left( {s + u} \right) = 6\left( 1 \right)\\ \Rightarrow 6s + 6u = 6\end{array}\]
Subtracting the equation \[6s + 6u = 6\] from the equation \[6s + 8u = - 4\], we get
\[\begin{array}{l} \Rightarrow 6s + 8u - \left( {6s + 6u} \right) = - 4 - 6\\ \Rightarrow 6s + 8u - 6s - 6u = - 10\end{array}\]
Subtracting the like terms, we get
\[ \Rightarrow 2u = - 10\]
Dividing both sides by 2, we get
\[ \Rightarrow u = - 5\]
Substituting \[u = - 5\] in the equation \[s + u = 1\], we get
\[ \Rightarrow s - 5 = 1\]
Adding 5 on both sides of the equation, we get
\[\begin{array}{l} \Rightarrow s - 5 + 5 = 1 + 5\\ \Rightarrow s = 6\end{array}\]
Substituting \[b = 1\], \[c = - 16\], \[d = - 4\], \[e = 48\], \[s = 6\], \[u = - 5\], \[t = 8\], and \[v = 6\] in the equation \[{x^4} + b{x^3} + c{x^2} + dx + e = \left( {{x^2} + sx + t} \right)\left( {{x^2} + ux + v} \right)\], we get
\[ \Rightarrow {x^4} + {x^3} - 16{x^2} - 4x + 48 = \left( {{x^2} + 6x + 8} \right)\left( {{x^2} - 5x + 6} \right)\]
We will factorise the two quadratic equations to find the roots of the quartic equation.
Factoring the quadratic equations using splitting the middle term, we get
\[ \Rightarrow {x^4} + {x^3} - 16{x^2} - 4x + 48 = \left( {{x^2} + 4x + 2x + 8} \right)\left( {{x^2} - 3x - 2x + 6} \right)\]
Factoring out the terms, we get
\[ \Rightarrow {x^4} + {x^3} - 16{x^2} - 4x + 48 = \left[ {x\left( {x + 4} \right) + 2\left( {x + 4} \right)} \right]\left[ {x\left( {x - 3} \right) - 2\left( {x - 3} \right)} \right]\]
Therefore, we get
\[ \Rightarrow {x^4} + {x^3} - 16{x^2} - 4x + 48 = \left( {x + 4} \right)\left( {x + 2} \right)\left( {x - 2} \right)\left( {x - 3} \right)\]
Since \[{x^4} + {x^3} - 16{x^2} - 4x + 48 = 0\], we get
\[ \Rightarrow \left( {x + 4} \right)\left( {x + 2} \right)\left( {x - 2} \right)\left( {x - 3} \right) = 0\]
Therefore, the roots of the quartic equation can be obtained by solving
\[x + 4 = 0\], \[x + 2 = 0\], \[x - 2 = 0\], and \[x - 3 = 0\]
Thus, we get
\[x = - 4,2, - 2,3\]
Thus, the roots of the quartic equation are 2, 3, \[ - 2\], and \[ - 4\].
Note:
We can verify our answer by substituting the roots obtained into the given equation and see if the answer is 0.
Substituting \[x = - 4\] in the equation \[{x^4} + {x^3} - 16{x^2} - 4x + 48 = 0\], we get
\[\begin{array}{l} \Rightarrow {\left( { - 4} \right)^4} + {\left( { - 4} \right)^3} - 16{\left( { - 4} \right)^2} - 4\left( { - 4} \right) + 48 = 256 - 64 - 16\left( {16} \right) + 16 + 48\\ \Rightarrow {\left( { - 4} \right)^4} + {\left( { - 4} \right)^3} - 16{\left( { - 4} \right)^2} - 4\left( { - 4} \right) + 48 = 256 - 64 - 256 + 16 + 48\\ \Rightarrow {\left( { - 4} \right)^4} + {\left( { - 4} \right)^3} - 16{\left( { - 4} \right)^2} - 4\left( { - 4} \right) + 48 = 0\end{array}\]
Substituting \[x = - 2\] in the equation \[{x^4} + {x^3} - 16{x^2} - 4x + 48 = 0\], we get
\[\begin{array}{l} \Rightarrow {\left( { - 2} \right)^4} + {\left( { - 2} \right)^3} - 16{\left( { - 2} \right)^2} - 4\left( { - 2} \right) + 48 = 16 - 8 - 16\left( 4 \right) + 8 + 48\\ \Rightarrow {\left( { - 2} \right)^4} + {\left( { - 2} \right)^3} - 16{\left( { - 2} \right)^2} - 4\left( { - 2} \right) + 48 = 16 - 8 - 64 + 8 + 48\\ \Rightarrow {\left( { - 2} \right)^4} + {\left( { - 2} \right)^3} - 16{\left( { - 2} \right)^2} - 4\left( { - 2} \right) + 48 = 0\end{array}\]
Substituting \[x = 2\] in the equation \[{x^4} + {x^3} - 16{x^2} - 4x + 48 = 0\], we get
\[\begin{array}{l} \Rightarrow {2^4} + {2^3} - 16{\left( 2 \right)^2} - 4\left( 2 \right) + 48 = 16 + 8 - 16\left( 4 \right) - 8 + 48\\ \Rightarrow {2^4} + {2^3} - 16{\left( 2 \right)^2} - 4\left( 2 \right) + 48 = 16 + 8 - 64 - 8 + 48\\ \Rightarrow {2^4} + {2^3} - 16{\left( 2 \right)^2} - 4\left( 2 \right) + 48 = 0\end{array}\]
Substituting \[x = 3\] in the equation \[{x^4} + {x^3} - 16{x^2} - 4x + 48 = 0\], we get
\[\begin{array}{l} \Rightarrow {3^4} + {3^3} - 16{\left( 3 \right)^2} - 4\left( 3 \right) + 48 = 81 + 27 - 16\left( 9 \right) - 12 + 48\\ \Rightarrow {3^4} + {3^3} - 16{\left( 3 \right)^2} - 4\left( 3 \right) + 48 = 81 + 27 - 144 - 12 + 48\\ \Rightarrow {3^4} + {3^3} - 16{\left( 3 \right)^2} - 4\left( 3 \right) + 48 = 0\end{array}\]
Therefore, we have verified that the roots of the quartic equation \[{x^4} + {x^3} - 16{x^2} - 4x + 48 = 0\] are 2, 3, \[ - 2\], and \[ - 4\].
The roots whose product is 6 are 2 and 3.
Here, we need to find the roots of the given quartic equation. We will find the product of the other roots of the quartic equation. Then, using Descartes’s solution, we will rewrite the quartic equation as a product of two quadratic equations. Finally, we will factorise the two quadratic equations to get the four roots of the given quartic equation.
Formula Used:
We will use the following formulas:
1) The product of all roots of a quartic equation of the form \[a{x^4} + b{x^3} + c{x^2} + dx + e = 0\] is given by \[\dfrac{e}{a}\].
2) According to Descartes’s solution, we can write a quartic equation \[a{x^4} + b{x^3} + c{x^2} + dx + e = 0\] as \[{x^4} + b{x^3} + c{x^2} + dx + e = \left( {{x^2} + sx + t} \right)\left( {{x^2} + ux + v} \right)\], which can be further simplified as \[\left( {{x^2} + sx + t} \right)\left( {{x^2} + ux + v} \right) = {x^4} + \left( {s + u} \right){x^3} + \left( {t + v + su} \right){x^2} + \left( {sv + tu} \right)x + tv\]. Here, \[b = s + u\], \[c = t + v + su\], \[d = sv + tu\], and \[e = tv\].
Complete step by step solution:
Let the four roots of the given quartic equation be \[\alpha \], \[\beta \], \[\gamma \], and \[\delta \].
It is given that the product of two roots is 6.
Thus, we can write
\[\alpha \beta = 6\]
Now, the product of all roots of a quartic equation of the form \[a{x^4} + b{x^3} + c{x^2} + dx + e = 0\] is given by \[\dfrac{e}{a}\].
Comparing the equation \[{x^4} + {x^3} - 16{x^2} - 4x + 48 = 0\] to the equation \[a{x^4} + b{x^3} + c{x^2} + dx + e = 0\], we get
\[a = 1\] and \[e = 48\]
Thus, we get the product of all roots of the quartic equation as
\[\dfrac{e}{a} = \dfrac{{48}}{1} = 48\]
The roots are \[\alpha \], \[\beta \], \[\gamma \], and \[\delta \].
Therefore, we can write
\[\alpha \beta \gamma \delta = 48\]
Substituting \[\alpha \beta = 6\] in the equation, we get
\[ \Rightarrow 6\gamma \delta = 48\]
Dividing both sides by 6, we get
\[ \Rightarrow \gamma \delta = 8\]
Using Descartes’s solution, we can factorise the quartic equation into two quadratic equations.
Comparing the coefficients of the terms in the equations \[{x^4} + {x^3} - 16{x^2} - 4x + 48 = 0\] and \[{x^4} + \left( {s + u} \right){x^3} + \left( {t + v + su} \right){x^2} + \left( {sv + tu} \right)x + tv\], we get
\[s + u = 1\]
\[sv + tu = - 4\]
\[tv = 48\]
The constant 48 is the product of 8 and 6.
Therefore, substituting \[t = 8\] and \[v = 6\], we get
\[6s + 8u = - 4\]
Multiplying both sides of the equation \[s + u = 1\] by 6, we get
\[\begin{array}{l} \Rightarrow 6\left( {s + u} \right) = 6\left( 1 \right)\\ \Rightarrow 6s + 6u = 6\end{array}\]
Subtracting the equation \[6s + 6u = 6\] from the equation \[6s + 8u = - 4\], we get
\[\begin{array}{l} \Rightarrow 6s + 8u - \left( {6s + 6u} \right) = - 4 - 6\\ \Rightarrow 6s + 8u - 6s - 6u = - 10\end{array}\]
Subtracting the like terms, we get
\[ \Rightarrow 2u = - 10\]
Dividing both sides by 2, we get
\[ \Rightarrow u = - 5\]
Substituting \[u = - 5\] in the equation \[s + u = 1\], we get
\[ \Rightarrow s - 5 = 1\]
Adding 5 on both sides of the equation, we get
\[\begin{array}{l} \Rightarrow s - 5 + 5 = 1 + 5\\ \Rightarrow s = 6\end{array}\]
Substituting \[b = 1\], \[c = - 16\], \[d = - 4\], \[e = 48\], \[s = 6\], \[u = - 5\], \[t = 8\], and \[v = 6\] in the equation \[{x^4} + b{x^3} + c{x^2} + dx + e = \left( {{x^2} + sx + t} \right)\left( {{x^2} + ux + v} \right)\], we get
\[ \Rightarrow {x^4} + {x^3} - 16{x^2} - 4x + 48 = \left( {{x^2} + 6x + 8} \right)\left( {{x^2} - 5x + 6} \right)\]
We will factorise the two quadratic equations to find the roots of the quartic equation.
Factoring the quadratic equations using splitting the middle term, we get
\[ \Rightarrow {x^4} + {x^3} - 16{x^2} - 4x + 48 = \left( {{x^2} + 4x + 2x + 8} \right)\left( {{x^2} - 3x - 2x + 6} \right)\]
Factoring out the terms, we get
\[ \Rightarrow {x^4} + {x^3} - 16{x^2} - 4x + 48 = \left[ {x\left( {x + 4} \right) + 2\left( {x + 4} \right)} \right]\left[ {x\left( {x - 3} \right) - 2\left( {x - 3} \right)} \right]\]
Therefore, we get
\[ \Rightarrow {x^4} + {x^3} - 16{x^2} - 4x + 48 = \left( {x + 4} \right)\left( {x + 2} \right)\left( {x - 2} \right)\left( {x - 3} \right)\]
Since \[{x^4} + {x^3} - 16{x^2} - 4x + 48 = 0\], we get
\[ \Rightarrow \left( {x + 4} \right)\left( {x + 2} \right)\left( {x - 2} \right)\left( {x - 3} \right) = 0\]
Therefore, the roots of the quartic equation can be obtained by solving
\[x + 4 = 0\], \[x + 2 = 0\], \[x - 2 = 0\], and \[x - 3 = 0\]
Thus, we get
\[x = - 4,2, - 2,3\]
Thus, the roots of the quartic equation are 2, 3, \[ - 2\], and \[ - 4\].
Note:
We can verify our answer by substituting the roots obtained into the given equation and see if the answer is 0.
Substituting \[x = - 4\] in the equation \[{x^4} + {x^3} - 16{x^2} - 4x + 48 = 0\], we get
\[\begin{array}{l} \Rightarrow {\left( { - 4} \right)^4} + {\left( { - 4} \right)^3} - 16{\left( { - 4} \right)^2} - 4\left( { - 4} \right) + 48 = 256 - 64 - 16\left( {16} \right) + 16 + 48\\ \Rightarrow {\left( { - 4} \right)^4} + {\left( { - 4} \right)^3} - 16{\left( { - 4} \right)^2} - 4\left( { - 4} \right) + 48 = 256 - 64 - 256 + 16 + 48\\ \Rightarrow {\left( { - 4} \right)^4} + {\left( { - 4} \right)^3} - 16{\left( { - 4} \right)^2} - 4\left( { - 4} \right) + 48 = 0\end{array}\]
Substituting \[x = - 2\] in the equation \[{x^4} + {x^3} - 16{x^2} - 4x + 48 = 0\], we get
\[\begin{array}{l} \Rightarrow {\left( { - 2} \right)^4} + {\left( { - 2} \right)^3} - 16{\left( { - 2} \right)^2} - 4\left( { - 2} \right) + 48 = 16 - 8 - 16\left( 4 \right) + 8 + 48\\ \Rightarrow {\left( { - 2} \right)^4} + {\left( { - 2} \right)^3} - 16{\left( { - 2} \right)^2} - 4\left( { - 2} \right) + 48 = 16 - 8 - 64 + 8 + 48\\ \Rightarrow {\left( { - 2} \right)^4} + {\left( { - 2} \right)^3} - 16{\left( { - 2} \right)^2} - 4\left( { - 2} \right) + 48 = 0\end{array}\]
Substituting \[x = 2\] in the equation \[{x^4} + {x^3} - 16{x^2} - 4x + 48 = 0\], we get
\[\begin{array}{l} \Rightarrow {2^4} + {2^3} - 16{\left( 2 \right)^2} - 4\left( 2 \right) + 48 = 16 + 8 - 16\left( 4 \right) - 8 + 48\\ \Rightarrow {2^4} + {2^3} - 16{\left( 2 \right)^2} - 4\left( 2 \right) + 48 = 16 + 8 - 64 - 8 + 48\\ \Rightarrow {2^4} + {2^3} - 16{\left( 2 \right)^2} - 4\left( 2 \right) + 48 = 0\end{array}\]
Substituting \[x = 3\] in the equation \[{x^4} + {x^3} - 16{x^2} - 4x + 48 = 0\], we get
\[\begin{array}{l} \Rightarrow {3^4} + {3^3} - 16{\left( 3 \right)^2} - 4\left( 3 \right) + 48 = 81 + 27 - 16\left( 9 \right) - 12 + 48\\ \Rightarrow {3^4} + {3^3} - 16{\left( 3 \right)^2} - 4\left( 3 \right) + 48 = 81 + 27 - 144 - 12 + 48\\ \Rightarrow {3^4} + {3^3} - 16{\left( 3 \right)^2} - 4\left( 3 \right) + 48 = 0\end{array}\]
Therefore, we have verified that the roots of the quartic equation \[{x^4} + {x^3} - 16{x^2} - 4x + 48 = 0\] are 2, 3, \[ - 2\], and \[ - 4\].
The roots whose product is 6 are 2 and 3.
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