Q. 52 F5.0( 1 Vote )

Solve the followi

Answer :


Let


We need to find the roots of Δ = 0.


Recall that the value of a determinant remains same if we apply the operation Ri Ri + kRj or Ci Ci + kCj.


Applying R2 R2 – R1, we get




Applying R3 R3 – R1, we get





Taking (b – x) and (c – x) common from R2 and R3, we get



Expanding the determinant along C1, we have


Δ = (b – x)(c – x)(1)[(1)(c2 + cx + x2) – (1)(b2 + bx + x2)]


Δ = (b – x)(c – x)[c2 + cx + x2 – b2 – bx – x2]


Δ = (b – x)(c – x)[c2 – b2 + cx – bx]


Δ = (b – x)(c – x)[(c – b)(c + b) + (c – b)x]


Δ = (b – x)(c – x)(c – b)(c + b + x)


The given equation is Δ = 0.


(b – x)(c – x)(c – b)(c + b + x) = 0


However, b ≠ c according to the given condition.


(b – x)(c – x)(c + b + x) = 0


Case – I:


b – x = 0 x = b


Case – II:


c – x = 0 x = c


Case – III:


c + b + x = 0 x = –(b + c)


Thus, b, c and –(b + c) are the roots of the given determinant equation.


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