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I have a chapter in my school course book on quadratic equations, in which we are learning how to solve nonquadratic-equations , by reducing them to quadratic form, the book describes 5 types of equations which can be solved by this method one of which is:

$$(x+a)(x+b)(x+c)(x+d)=k$$

According to the book to solve these kind of equations, we have to find which two pair of the above constants, which have the same sum eg $a+b=c+d$ Now we should rearrange the terms in the equation so as to get terms with one of the pair of constants adjacent to each other. eg if $a+c=b+d$ then we would rearrange as $$(x+a)(x+c)(x+b)(x+d)=k$$

Now after this step we would multiply the terms, constants of which form the pairs, with each other eg in above case $$(x^2+cx+ax+ac)(x^2+bx+dx+bd)=k$$

Now as we know $a+c=b+d$ we can place a variable say $y=x^2+cx+ax=x^2+bx+dx$ into the above equation and get $$(y+ac)(y+bd)=k$$ which could then be converted to a quadratic equation and then be solved as one, after which we could plug in value of $y$ and get the values of $x$ by applying an appropriate method. That being said, the book didn't give examples of the equations of the form $$(x+a)(ex+b)(ex+c)(x+d)=k$$ and has asked us to solve such type of an equation in the excercise, and i dont have a clue as how can i solve it, as the method of the book, described above cannot be applied to it. ANY HELP WOULD BE MUCH APPRECIATED

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  • $\begingroup$ Hi and welcome to the site. I helped you with typesetting your question. You can learn to do this on your own with LaTeX and MathJax and also see how others have typeset stuff. $\endgroup$ Commented Oct 4, 2015 at 13:22
  • $\begingroup$ As @mathreadler said, use $MathJax$ while posting. $\endgroup$
    – user249332
    Commented Oct 4, 2015 at 13:24
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    $\begingroup$ If you write down the particular equation you need to solve, probably solutions will be posted quickly. Probably what you are expected to use is a mild variant of the idea described earlier. $\endgroup$ Commented Oct 4, 2015 at 13:28

1 Answer 1

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Hint:

since $e\ne 0$:

$$(x+a)(ex+b)(ex+c)(x+d)=k \iff$$ $$ e^2\left(x+a\right)\left(x+\frac{b}{e}\right)\left(x+\frac{c}{e}\right)\left(x+d\right)=k \iff$$ $$ \left(x+a\right)\left(x+\frac{b}{e}\right)\left(x+\frac{c}{e}\right)\left(x+d\right)=\frac{k}{e^2} $$

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