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question mark

find all a (a is any real number) for wich the equation a^3+ a^2|a+x| + |a^2x+1| = 1 has not less than four different solutions wich are integers

Shubham Agrawal , 13 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello Student,
Please find answer to your question below

$a^3 + a^2|a+x| + |a^2x+1| = 1$
$|a+x| = a+x; x\geq -a$
$|a+x| = -a-x; x\leq -a$
$|a^2x+1| = a^2x+1;x\geq \frac{-1}{a^2}$
$|a^2x+1| = -a^2x-1;x\leq \frac{-1}{a^2}$
For a > 0
$-a\leq x\leq \frac{-1}{a^2}$
$a^3 + a^2|a+x| + |a^2x+1| = 1$
$a^3 + a^2(a+x) - (a^2x+1) = 1$
$2a^3 = 2$
$a^3 = 1$
$\Rightarrow a = 1$
$x\leq -a$
$a^3 + a^2|a+x| + |a^2x+1| = 1$
$a^3 - a^2(a+x) - (a^2x+1) = 1$
$-2a^{2}x = 2$
$-a^{2}x = 1$
$a^{2} = \frac{-1}{x}$
$a = \sqrt{\frac{-1}{x}}$
$x\geq \frac{-1}{a^2}$
$a^3 + a^2|a+x| + |a^2x+1| = 1$
$a^3 + a^2(a+x) + (a^2x+1) = 1$
$a^3 + a^2x = 0$
$a^2(a + x) = 0$
$a = 0$
$a = -x;x\leq 0$

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