Question

If x2 – 15 |x| + 26 = 0, where x is real, what is the minimum possible value of x?

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{x^2-15|x|+26=0}$

$\sf{\implies\,|x|^2-15|x|+26=0}$

$\sf{\implies\,|x|^2-13|x|-2|x|+26=0}$

$\sf{\implies|x|(|x|-13)-2(|x|-13)=0}$

$\sf{\implies(|x|-2)(|x|-13)=0}$

$\sf{\implies\,|x|=2\,\,\,\,\,or\,\,\,\,\,|x|=13}$

$\sf{\implies\,x=2\,\,\,\,\,or\,\,\,\,\,x=-2\,\,\,\,\,or\,\,\,\,\,x=13\,\,\,\,\,or\,\,\,\,\,x=-13}$

So, minimum possible value of x is -13