Let α, β be the roots of the equation x²-√2 x+√6=0 and 1/α²+1, 1/β²+1 be the roots of the equation x²+a x+b=0. Then the roots of the equation x²-(a+b-2) x+(a+b+2)=0 are :

Solution

Maths sir · 5 steps

You only need a+b. Put x = 1 into x² + ax + b and it comes out in one line.

1. Use x equals 1. If r₁, r₂ are the roots of x² + ax + b = 0, then x² + ax + b = (x - r₁)(x - r₂). Putting x = 1 gives 1 + a + b = (1 - r₁)(1 - r₂). Here 1 - r₁ = -1/α² and 1 - r₂ = -1/β².

   1 + a + b = (-1/α²)(-1/β²) = 1/(αβ)² = 1/(√6)² = 1/6

2. Find a plus b. From the first equation, αβ = √6, which is the value used above. So a + b = 1/6 - 1 = -5/6.
3. Build the new equation. a + b - 2 = -17/6 and a + b + 2 = 7/6. The equation becomes x² + 17/6x + 7/6 = 0, that is, 6x² + 17x + 7 = 0.
4. Solve it. The discriminant is 17² - 4 × 6 × 7 = 289 - 168 = 121 > 0, so the roots are real. They are x = (-17 ± 11)/12, which gives -1/2 and -7/3.
5. So. Both roots are real and negative. Their sum -17/6 is negative and their product 7/6 is positive, which confirms it. The answer is (b).

The idea to keep. If a question needs 1 + a + b for x² + ax + b, that is the quadratic at x = 1, which equals (1 - r₁)(1 - r₂). Use this before you expand any sums and products of roots.

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