If the value of real number a>0 for which x²-5 a x+1=0 and x²-a x-5=0
have a common real root is 3/√(2 β) then β is equal to ___________.

Solution

Maths sir · 4 steps

Subtract the two equations: the x² terms vanish and the common root comes out in terms of a.

1. Remove the square. Let α be the common root, so it satisfies both equations. Subtract the second from the first: (α² - 5aα + 1) - (α² - aα - 5) = 0, which gives -4aα + 6 = 0.
2. The common root. So α = 3/2a. This is real for every a > 0.
3. Put it back. Substitute into x² - ax - 5 = 0: 9/4a² - 3/2 - 5 = 0, so 9/4a² = 13/2 and a² = 9/26. The first equation checks out too: 13/2 - 15/2 + 1 = 0.

   a = 3/√26 = 3/√(2 × 13)

4. So. Matching with 3/√2β gives 2β = 26, so β = 13.

The common slip: 26. Reading a = 3/√26 and taking β as 26 forgets the factor 2 inside √2β; it is 2β that equals 26.

The idea to keep. For a common root of two quadratics with the same x² coefficient, subtract them. You get a linear equation for the root, then substitute it back into either equation.

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