If p- q- r are three distinct real numbers- p- 0 such that x 2 - qx -pr-0 and x 2 - rx -pq-0 have a common root- then the value of p-q-r is

Let α be the common root for both equations then α^2 + qα + pr = 0…(1) α^2 + rα + pq = 0…(2) Subtracting (2) from (1) we get : (q ...

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Standard XII

Mathematics

Higher Order Equations

If p, q, r ...

Question

If $p, q, r$ are three distinct real numbers, $(p \neq 0)$ such that $x^{2} + q x + p r = 0$ and $x^{2} + r x + p q = 0$ have a common root, then the value of $p + q + r$ is

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Solution

Let $α$ be the common root for both equations then
$α^{2} + q α + p r = 0 \dots (1)$
$α^{2} + r α + p q = 0 \dots (2)$
Subtracting $(2)$ from $(1)$ we get :
$(q - r) α + p (r - q) = 0$
$α = p$
Substituting value of $α = p$ in equation $(1)$ we get
$p^{2} + p q + p r = 0$
Hence, $p + q + r = 0$ .

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If p,q,r are three distinct real numbers, (p≠0) such that x2+qx+pr=0 and x2+rx+pq=0 have a common root, then the value of p+q+r is

Let α be the common root for both equations thenα2+qα+pr=0…(1)α2+rα+pq=0…(2)Subtracting (2) from (1) we get :(q−r)α+p(r−q)=0α=pSubstituting value of α=p in equation (1) we get p2+pq+pr=0Hence, p+q+r=0.

If $p, q, r$ are three distinct real numbers, $(p \neq 0)$ such that $x^{2} + q x + p r = 0$ and $x^{2} + r x + p q = 0$ have a common root, then the value of $p + q + r$ is