anemone said:Solve for all real solutions for the system below:
$(x^2+3x+2)(x^2-2x-1)(x^2-7x+12)+24=0$
greg1313 said:\(\displaystyle (x^2+3x+2)(x^2-2x-1)(x^2-7x+12)+24\)
\(\displaystyle =x^6-6x^5+40x^3-13x^2-70x\)
\(\displaystyle =x(x^5-6x^4+40x^2-13x-70)\)
\(\displaystyle =x(x-2)(x^4-4x^3-8x^2+24x+35)\)
\(\displaystyle =x(x-2)(x^2-2x-5)(x^2-2x-7)=0\)
\(\displaystyle \implies x\in\left\{0,2,1\pm\sqrt6,1\pm2\sqrt2\right\}\)
When solving an equation or problem, finding the real solution(s) means finding the value(s) of the variable(s) that make the equation true. In other words, it is finding the solution(s) that actually exist and have a real, tangible meaning.
A real solution is a value of the variable that makes the equation true when substituted into the equation. This can be checked by plugging in the value and simplifying both sides of the equation to see if they are equal.
A real solution is a value of the variable that results in a real number when substituted into the equation. An imaginary solution is a value that results in an imaginary number, typically denoted by i, when substituted into the equation. Imaginary solutions often occur when solving equations with complex numbers or when taking the square root of a negative number.
Yes, an equation can have multiple real solutions. This means that there are multiple values of the variable that make the equation true. These solutions can be distinct or they can be repeated, depending on the equation and the type of problem being solved.
To solve for real solutions, you need to isolate the variable on one side of the equation and simplify the other side. This may involve using algebraic techniques such as factoring, combining like terms, or using the quadratic formula. Once the variable is isolated, you can plug in any potential solutions and check if they make the equation true.