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Question: Let a, b > 0 and $\Delta = \begin{vmatrix} -x & a & b \\ b & -x & a \\ a & b & -x \end{vmatrix}$, th...

Let a, b > 0 and Δ=∣−xabb−xaab−x∣\Delta = \begin{vmatrix} -x & a & b \\ b & -x & a \\ a & b & -x \end{vmatrix}, then

A

a + b - x is a factor of Δ\Delta

B

x2+(a+b)x+a2+b2−abx^2 + (a+b)x + a^2 + b^2 - ab is a factor of Δ\Delta

C

Δ\Delta = 0 has three real roots if a = b

D

a + b + x is a factor of Δ\Delta

Answer

Options (A), (B), and (C) are correct.

Explanation

Solution

Solution:

We are given

Δ=∣−xabb−xaab−x∣.\Delta=\begin{vmatrix} -x & a & b \\ b & -x & a \\ a & b & -x \end{vmatrix}.

Step 1. Expand the Determinant

Expand along the first row:

Δ=(−x)∣−xab−x∣−a∣baa−x∣+b∣b−xab∣=(−x)[(−x)(−x)−a⋅b]−a[b(−x)−a⋅a]+b[b⋅b−(−x)⋅a]=(−x)[x2−ab]−a[−bx−a2]+b[b2+ax]=−x3+abx+abx+a3+b3+abx=−x3+3ab x+a3+b3.\begin{aligned} \Delta &= (-x)\begin{vmatrix} -x & a \\ b & -x \end{vmatrix} - a\begin{vmatrix} b & a \\ a & -x \end{vmatrix} + b\begin{vmatrix} b & -x \\ a & b \end{vmatrix}\\[1mm] &= (-x)\Big[(-x)(-x)-a\cdot b\Big] - a\Big[b(-x)-a\cdot a\Big] + b\Big[b\cdot b-(-x)\cdot a\Big]\\[1mm] &= (-x)\Big[x^2-ab\Big] - a\Big[-bx-a^2\Big] + b\Big[b^2+ax\Big]\\[1mm] &= -x^3+abx + abx+a^3 + b^3+abx\\[1mm] &= -x^3+3ab\,x+a^3+b^3. \end{aligned}

Step 2. Factorize the Cubic

Notice that

a3+b3=(a+b)(a2−ab+b2).a^3+b^3 = (a+b)(a^2-ab+b^2).

Test x=a+bx = a+b:

Δ(x=a+b)=−(a+b)3+3ab(a+b)+a3+b3.\Delta(x=a+b) = - (a+b)^3 + 3ab(a+b)+a^3+b^3.

Since

(a+b)3=a3+3a2b+3ab2+b3,(a+b)^3=a^3+3a^2b+3ab^2+b^3,

we have:

Δ(x=a+b)=−[a3+3a2b+3ab2+b3]+3ab(a+b)+a3+b3=0.\Delta(x=a+b) = -\left[a^3+3a^2b+3ab^2+b^3\right] + 3ab(a+b)+a^3+b^3=0.

Thus, (a+b−x)(a+b-x) is a factor. Dividing, we obtain:

Δ=(a+b−x)[x2+(a+b)x+(a2−ab+b2)].\Delta = (a+b-x)\left[x^2+(a+b)x+(a^2-ab+b^2)\right].

Step 3. Analyze the Options

  • Option (A): Since a+b−xa+b-x is a factor, (A) is correct.
  • Option (B): The quadratic factor is x2+(a+b)x+(a2−ab+b2)x^2+(a+b)x+(a^2-ab+b^2) which matches the given expression, so (B) is correct.
  • Option (C): For a=ba=b (say a=b=c>0a=b=c>0), the factors become: a+b−x=2c−xandx2+2cx+(c2−c2+c2)=x2+2cx+c2=(x+c)2.a+b-x=2c-x \quad \text{and} \quad x^2+2cx+(c^2-c^2+c^2)=x^2+2cx+c^2=(x+c)^2. Thus, Δ=(2c−x)(x+c)2,\Delta=(2c-x)(x+c)^2, giving roots x=2cx=2c and x=−cx=-c (a repeated root). All roots are real. Hence, (C) is correct.
  • Option (D): Testing x=−a−bx=-a-b does not yield zero in general, so (D) is false.

Final Answer: Options (A), (B), and (C) are correct.

Core Explanation:

  1. Compute Δ=−x3+3abx+a3+b3\Delta = -x^3+3abx+a^3+b^3.
  2. Factorize as Δ=(a+b−x)[x2+(a+b)x+(a2−ab+b2)]\Delta = (a+b-x)\left[x^2+(a+b)x+(a^2-ab+b^2)\right].
  3. For a=ba=b: Δ=(2a−x)(x+a)2\Delta = (2a-x)(x+a)^2, hence three real roots.