Grade 11Algebra

if there is a common root of x2+px+q=0and x2+qx+pr=0,then proof that p+q+r=0

Debasish Bhattacharya

7 Years agoGrade 11

7 Years ago

The equations given are slightly wrong. By solving them we get wrong result.

Equations should be x² + px + qr = 0 and x² + qx + pr = 0.

Concept : Let a₁x² + b₁ x + c₁ = 0 and a₂x² + b₂ x + c₂ = 0.

Let $\alpha$ be the common root of the above equations.

By cross multiplication method, we get

$\alpha$ ² / (b₁c₂ – b₂c₁) = $\alpha$ / (c₁ a₂ – c₂ a₁) = 1 / (a₁b₂ – a₂ b₁)

$\therefore$ $\alpha$ = (b₁c₂ – b₂c₁) / (c₁ a₂ – c₂ a₁) = (c₁ a₂ – c₂ a₁) / (a₁b₂ – a₂ b₁) ...(1)

Consider (b₁c₂ – b₂c₁) / (c₁ a₂ – c₂ a₁) = (c₁ a₂ – c₂ a₁) / (a₁b₂ – a₂ b₁)

$\Rightarrow$ (b₁c₂ – b₂c₁)(a₁b₂ – a₂ b₁) = (c₁ a₂ – c₂ a₁)² …(2), which is the condition for common root.

Comparing given equations x² + px + qr = 0 and x² + qx + pr = 0 with

a₁x² + b₁ x + c₁ = 0 and a₂x² + b₂ x + c₂ = 0,

we get a₁= 1, b₁ = p, c₁ = qr, a₂ = 1, b₂ = q, c₂ = pr.

Putting these values in (2), we get

(p²r – q²r)(q – p) = (qr – pr)² i.e. r(p² – q²)(q – p) = r²(q – p)² ,now r(q – p) will get cancelled from both sides

(p² – q²) = r(q – p) or (p + q)(p – q) = r(q – p)

$\therefore$ (p + q) = – r $\Rightarrow$ p + q + r = 0.

(Note : You can memorise result (2) to solve these type of questions.)