Math, asked by smithi4167, 1 year ago

prove that equations (q-r)x^2+(r-p)x+p-q=0 and (r-p)x^2+(p-q)x+q-r=0 have a common root

Answers

Answered by pinquancaro
4

We have to prove that the equations (q-r)x^2+(r-p)x+(p-q)=0 and (r-p)x^2+(p-q)x+(q-r)=0 have a common root.

Let \alpha be the common root of the given equations.

So, (q-r)\alpha^2+(r-p)\alpha+(p-q)=0 and (r-p)\alpha^2+(p-q)\alpha+(q-r)=0

Solving these two equations by cross multiplication method, we get

\frac{\alpha^2}{(q^2-r^2)(q^2-r^2)-(r^2-p^2)(r^2-p^2)} = \frac{\alpha}{(r^2-p^2)(p^2-q^2)-(q^2-r^2)(p^2-q^2)} = \frac{1}{(r^2-p^2)(p^2-q^2)-(q^2-r^2)(p^2-q^2)}

By comparing the last two parts of the above equation, we get

\frac{\alpha}{(r^2-p^2)(p^2-q^2)-(q^2-r^2)(p^2-q^2)} = \frac{1}{(r^2-p^2)(p^2-q^2)-(q^2-r^2)(p^2-q^2)}

\implies \frac{\alpha}{(p^2-q^2)((r^2-p^2)-(q^2-r^2))}=\frac{1}{(p^2-q^2)((r^2-p^2)-(q^2-r^2))}

\implies \frac{\alpha}{(p^2-q^2)(2r^2-p^2-q^2)} =\frac{1}{(p^2-q^2)(2r^2-p^2-q^2)}

\implies \alpha=1

Hence, the given equations have a common root.

Answered by junejaabhilasha
2

Answer:

Step-by-step explanation:

Solution is in the attachment provided below

Hope it helps!

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