Question

Prove that equations (q-r)x2 + (r-p)x + p -q=0 and (r-p)x2 + (p-q)x + q-r=0 have a common root

Answer 1

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.

Answer 2

Answer:

Step-by-step explanation:

Solution is in the attachment provided below.

Hope it helps !