Question

समीकरण (q-r) x2 + (r - p) x + (p-q) = 0 के मूल हैं-​

Answer 1

Answer:

Consider the given equations:

(q-r)x^2+(r-p)x+(p-q)=0 and (r-p)x^2+(p-q)x+(q-r)=0

Comparing these equations to the general form of the quadratic equation,

we get a_1 =(q-r) , b_1=(r-p) ,c_1= (p-q)

a_2 =(r-p) , b_2=(p-q) ,c_2= (q-r)

The equations have a common root when

(c_1a_2-c_2a_1)^2 = (b_1c_2-b_2c_1)(a_1b_2-a_2b_1)

Consider   (c_1a_2-c_2a_1)^2

= [(p-q)(r-p)-(q-r)(q-r)]^2

= (pr-p^2-qr+qp-q^2-r^2+2qr)^2

= (pr-p^2-q^2+qp-r^2+qr)^2

=(-p^2-q^2-r^2+pq+pr+rq)^2

Consider (b_1c_2-b_2c_1)(a_1b_2-a_2b_1)

= [(r-p)(q-r)-(p-q)(p-q)] [(q-r)(p-q) - (r-p)(r-p)]

= (-p^2-q^2-r^2+pq+pr+rq)(-p^2-q^2-r^2+pq+pr+rq)

= (-p^2-q^2-r^2+pq+pr+rq)^2

Therefore, (c_1a_2-c_2a_1)^2 = (b_1c_2-b_2c_1)(a_1b_2-a_2b_1)

Hence, the given equations have a common root.

Step-by-step explanation:

now can you understand :)