- #1
kaliprasad
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solve the following system of equations in real a,b,c,d
$a+ b =8 $
$ab + c +d = 23 $
$ad + bc = 28$
$cd = 12$
$a+ b =8 $
$ab + c +d = 23 $
$ad + bc = 28$
$cd = 12$
Solve System of Equations: a+b, ab+c+d, ad+bc, cd
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In summary, a system of equations is a set of equations with common variables that need to be solved simultaneously. The most common methods for solving a system of equations are substitution and elimination, but there are also advanced methods like Gaussian elimination and Cramer's rule. An example of solving a system of equations involves substituting values from one equation into the others to find a unique solution. A system of equations can have one solution, no solutions, or infinitely many solutions depending on the consistency and dependence of the equations. Real-life applications of solving systems of equations can be found in fields such as engineering, economics, physics, and chemistry for tasks like designing structures, modeling supply and demand, predicting motion, and analyzing chemical reactions.
- #2
kaliprasad
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No answer yet
hint
hint
form a quartic equation
- #3
anemone
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kaliprasad said:
kaliprasad said:
My solution:
If we form a quartic equation as the product of two quadratic equations as follow, we have:
$\begin{align*}(x^2+ax+c)(x^2+bx+d)&=x^4+(a+b)x^3+(ab+c+d)x^2+(ad+bc)x+cd\\&=x^4+8x^3+23x^2+28x+12\\&\overset{I}{=}((x+1)(x+2))((x+2)(x+3))=(x^2+3x+2)(x^2+5x+6)\\&\overset{II}{=}((x+1)(x+3))((x+2)(x+2))=(x^2+4x+3)(x^2+4x+4)\\&\overset{III}{=}((x+2)(x+3))((x+1)(x+2))=(x^2+5x+6)(x^2+3x+2)\\&\overset{IV}{=}((x+2)(x+2))((x+1)(x+3))=(x^2+4x+4)(x^2+4x+3)\end{align*}$
Hence, $(a,\,b,\,c,\,d)=(3,\,5,\,2,\,6),\,(4,\,4,\,3,\,4),\,(5,\,3,\,6,\,2),\,(4,\,4,\,4,\,3)$.
$\begin{align*}(x^2+ax+c)(x^2+bx+d)&=x^4+(a+b)x^3+(ab+c+d)x^2+(ad+bc)x+cd\\&=x^4+8x^3+23x^2+28x+12\\&\overset{I}{=}((x+1)(x+2))((x+2)(x+3))=(x^2+3x+2)(x^2+5x+6)\\&\overset{II}{=}((x+1)(x+3))((x+2)(x+2))=(x^2+4x+3)(x^2+4x+4)\\&\overset{III}{=}((x+2)(x+3))((x+1)(x+2))=(x^2+5x+6)(x^2+3x+2)\\&\overset{IV}{=}((x+2)(x+2))((x+1)(x+3))=(x^2+4x+4)(x^2+4x+3)\end{align*}$
Hence, $(a,\,b,\,c,\,d)=(3,\,5,\,2,\,6),\,(4,\,4,\,3,\,4),\,(5,\,3,\,6,\,2),\,(4,\,4,\,4,\,3)$.
But...I'm not proud of myself for this solution because...
Without the hint, I could have never solved this great challenge!
Thank you so much, kaliprasad for the hint and also for sharing with us of this great challenge! I could tell I've fallen in love with this problem at first sight!Hahaha...
Thank you so much, kaliprasad for the hint and also for sharing with us of this great challenge! I could tell I've fallen in love with this problem at first sight!Hahaha...
Last edited:
- #4
kaliprasad
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anemone said:My solution:
If we form a quartic equation as the product of two quadratic equation as follow, we have:
$\begin{align*}(x^2+ax+c)(x^2+bx+d)&=x^4+(a+b)x^3+(ab+c+d)x^2+(ad+bc)x+cd\\&=x^4+8x^3+23x^2+28x+12\\&\overset{I}{=}((x+1)(x+2))((x+2)(x+3))=(x^2+3x+2)(x^2+5x+6)\\&\overset{II}{=}((x+1)(x+3))((x+2)(x+2))=(x^2+4x+3)(x^2+4x+4)\\&\overset{III}{=}((x+2)(x+3))((x+1)(x+2))=(x^2+5x+6)(x^2+3x+2)\\&\overset{IV}{=}((x+2)(x+2))((x+1)(x+3))=(x^2+4x+4)(x^2+4x+3)\end{align*}$
Hence, $(a,\,b,\,c,\,d)=(3,\,5,\,2,\,6),\,(4,\,4,\,3,\,4),\,(5,\,3,\,6,\,2),\,(4,\,4,\,4,\,3)$.
But...I'm not proud of myself for this solution because...
Without the hint, I could have never solved this great challenge!
Nothing more to write to make it right(pun intended)
Hence it is closed
- #5
CellCoree
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To solve this system of equations, we can use substitution or elimination methods. Let's start by solving for c and d in terms of a and b in the third equation, $ad + bc = 28$. We can rewrite this as $c = \frac{28 - ad}{b}$. Similarly, we can rewrite the fourth equation, $cd = 12$, as $d = \frac{12}{c}$. We can then substitute these expressions into the second equation, $ab + c + d = 23$, to get $ab + \frac{28 - ad}{b} + \frac{12}{c} = 23$. Multiplying both sides by bc, we get $abc + 28c - ad^2 = 23bc$. We can rearrange this to get $ad^2 + abc - 23bc = 28c$. This gives us a quadratic equation in terms of d, which we can solve using the quadratic formula. Once we have found the value of d, we can plug it back into our expression for c and solve for c. Then, we can plug in the values of c and d into the first equation, $a + b = 8$, to solve for a and b. This will give us the values of a, b, c, and d that satisfy all four equations in the system.
Related to Solve System of Equations: a+b, ab+c+d, ad+bc, cd
1. Can you explain what a system of equations is?
A system of equations is a set of equations that are related to each other and have common variables. The goal is to find values for the variables that satisfy all of the equations in the system.
2. How do I solve a system of equations?
The most common method for solving a system of equations is by substitution. This involves solving one equation for a variable and then substituting that value into the other equations. Another method is elimination, where you add or subtract equations to eliminate a variable. There are also advanced methods like Gaussian elimination and Cramer's rule.
3. Can you give an example of solving a system of equations?
Sure, let's say we have the system of equations a+b=5, ab+c+d=10, ad+bc=8, and cd=3. First, we can solve for a in the first equation to get a=5-b. Then, we can substitute that into the second equation to get (5-b)b+c+d=10. We can simplify this to -b^2+c+d-5b+10=0. Similarly, we can solve for d and substitute into the third equation to get (5-b)(c+b)=8. Finally, we can substitute these values into the fourth equation to get (5-b)(c+b)b=3. We can then solve for b, and use that value to find the other variables.
4. Can a system of equations have more than one solution?
Yes, a system of equations can have one solution, no solutions, or infinitely many solutions. If the equations are inconsistent, meaning they contradict each other, there will be no solutions. If the equations are dependent, meaning one equation is a multiple of another, there will be infinitely many solutions. Otherwise, there will be one unique solution.
5. What are some real-life applications of solving systems of equations?
Solving systems of equations is used in many fields such as engineering, economics, physics, and chemistry. For example, in engineering, systems of equations can be used to determine the optimal design for a structure. In economics, systems of equations can be used to model supply and demand in a market. In physics, they can be used to predict the motion of objects. In chemistry, they can be used to analyze chemical reactions.
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