How many distinct real roots does (x^2+x-2)^3+(8-2x^2)^3=(x^2+3x+2)^3 have?

Bruce Rosario

Bruce Rosario

Answered question

2022-04-23

How many distinct real roots does
(x2+x2)3+(82x2)3=(x2+3x+2)3 have?

Answer & Explanation

2sze1c1se3nh

2sze1c1se3nh

Beginner2022-04-24Added 17 answers

Using the idea in Aryabhata's comment, since
(82x2)3=23(4x2)3=8(2x)3(2+x)3
your second equation
(2+6x2)(x+2)3=(82x2)3
is equivalent to
(2+6x2)(x+2)3=8(2x)3(2+x)3
So the factor (x+2)3 in both sides yields the real triple root x=2. For x2 this equation (1) is equivalent to
(2+6x2)=8(x2)3
(2+6x2)+8(x2)3=0
(1+3x2)+4(x2)3=0
By inspection we see that x=1 is a root and by long division or Ruffini's rule, we find:
(1+3x2)+4(x2)3=4x321x2+48x31
=(x1)(4x217x+31)
Alternatively we could have applied the rational zero theorem: all the rational roots of the equation
dnxn+dn1xn1++d0=0
where all the coefficients are integers, are of the form pq=a factor of  d0a factor of  dn
Hence, for x2, the equation is equivalent to
(x1)(4x217x+31)=0
Since the discriminant of the quadratic term is negative
=1724×4×31=207<0
the original equation has the real solution x=1 due to the factor (x1) and the triple reals solution x=-2 due to the common factor (x+2)3. In the total, two distinct real roots.

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