The number of distinct real roots of the equation x⁵(x³-x²-x+1)+x(3 x³-4 x²-2 x+4)-1=0 is ______________.

Solution

Maths sir · 4 steps

Spot that 3x⁴-4x³-2x²+4x-1 is (3x-1)(x³-x²-x+1), and the whole equation factorises.

1. Find the common factor. Multiply out (3x-1)(x³-x²-x+1). You get 3x⁴-3x³-3x²+3x-x³+x²+x-1 = 3x⁴-4x³-2x²+4x-1. That is exactly the bracket after x in the second part, so x⁵ and 3x-1 both multiply the same cubic.

   (x⁵+3x-1)(x³-x²-x+1)=0 ⇒ (x⁵+3x-1)(x-1)²(x+1)=0

2. The cubic factor. x³-x²-x+1 = x²(x-1)-(x-1) = (x-1)²(x+1). It gives the roots x=1 (repeated) and x=-1, which are two distinct values.
3. The quintic factor. Let f(x)=x⁵+3x-1. Then f'(x)=5x⁴+3>0 for every x, so f is strictly increasing and has exactly one real root. Since f(0)=-1<0 and f(1)=3>0, this root lies in (0,1), so it is neither 1 nor -1.
4. So. The distinct real roots are 1, -1 and one root in (0,1). The answer is 3.

The common slip: 4. Counting the double root x=1 twice gives 4, but the question asks for distinct roots, so it counts once.

The idea to keep. When a big polynomial looks random, try to make one bracket appear twice. Then for the leftover factor, check the sign of the derivative: if it never changes sign, there is exactly one real root.

Answer from the NTA final answer key. Solution by PrepWiser.