Question

 Let a, b, c be real numbers a≠0. If p is a root of a2y2+by + c = 0, q is the root of a2y2+by + c = 0 and 0 <p< q, then the equation a2y2+by + c = 0 has a root r that always satisfies

(a) r = (p+q)/2
(b) r = p + (q/2)
(c) r = a
(d) p <r <q

​

Answer 1

Step-by-step explanation:

Take f(x)=a

2

x

2

+2bx+2c

Now, f(α)=2(a

2

α

2

+bα+c)−a

2

α

2

……(1)

Since α is a root of a

2

x

2

+bx+c

So (1) becomes −a

2

α

2

Then f(β)=−2(a

2

β

2

−bβ−c)+4a

2

β

2

……(2) which similarly becomes f(β)=2a

2

β

2

Since f(α)f(β)=−4α

2

β

2

a

4

<0

Hence the root of f(x) ie γ lies between α and β

i.e. α<γ<β.

Answer 2

Step-by-step explanation:

$p \leqslant r \leqslant q$

Answer is option d