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Given that

$x^4= (x-c)^2$

which c is a real constant number. If the above solution has four real roots. Hence, c must be

a. $$-\frac{1}{4}\le c \le \frac{1}{4}$$

b.$$c \le -\frac{1}{4}$$

c.$$ c \le\frac{1}{4}$$

d.$$c \ge\frac{1}{4}$$

e. any number for c

so i use this kind of method, but i dont know what errors that i make.

let us root both sides

$\sqrt{x^4}= \sqrt{(x-c)^2}$

$\rightarrow$ $x^2= (x-c)$

then i use discriminant method (which D $\gt 0$) Unfortunately, my answer only gives me: $$ c \le\frac{1}{4}$$

what did i miss?

Iqbal Jurist
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  • 1
    Note that the above is equivalent to the expression $x^2=+(x-c)\lor x^2=-(x-c)$. When do these equation both have two real roots each? Check the discriminant – b00n heT May 26 '16 at 11:44
  • hint you can rewrite $x^4-(x-c)^2 = 0$ and then factor using $a^2-b^2 = (a+b)(a-b)$ – mathreadler May 26 '16 at 11:48

2 Answers2

1

You have $x^2=\pm(x-c)$. So either $x^2+x-c=0$ or $x^2-x+c=0$. The first has real roots iff $1+4c\ge0$ and the second iff $1-4c\ge0$. So for all roots real we need $-\frac{1}{4}\le c\le\frac{1}{4}$.

almagest
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We can rewrite it by factorization. x^4-(x-c)^2=(x^2-x+c)(x^2+x-c), if each factor of it has 2 roots, then it may have 4 roots. Use discriminant to get the result! 1^2-4c>=0 and 1^2+4c>=0 give (a).

tanghe2014
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