Example 4 Given that a is an integer, the equation concerning x x2+1x2−x2+14∣x∣+2−a=0
has real roots. Then the possible values of a are (2008, I Love Mathematics Junior High School Summer Camp Mathematics Competition)
A number or a short expression. Spacing and $ signs are ignored.
Solution
【Analysis】Take any real root x0 of the equation. Then x02+1x02−x02+14∣x0∣+2=a.
Notice that x02+1x02=(x02+1∣x0∣)2. Let x02+1∣x0∣=t. By x02+1>x02=∣x0∣ ⇒x02+1∣x0∣<1⇒0⩽t<1 ⇒a=t2−4t+2=(t−2)2−2. Also 0⩽t<1⇒−1<(t−2)2−2⩽2 ⇒−1<a⩽2⇒a's possible values are 0,1,2.
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