If the roots of quadratic equation
are in ratio p:q , show that
where a and c are real number .
**Please Answer It.....(Don't Spam) & **Irrilivant Answers**
Answers
Answer:
hlo dear
Good evening
god bless you
and mark me as brain list
i hope u understand dear.......
Step-by-step explanation:
: \sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{c}{a}}=0
q
p
+
p
q
+
a
c
=0
Given quadratic equation,
ax^{2}+cx+c=0ax
2
+cx+c=0
Roots are in the ratio p:q.
Let the roots be pk, qk.
sum of the roots = -c/a
pk + qk = -c/a
p + q = (-c/a)(1/k) --------------(1)
product of roots = c/a
pk.qk = c/a
pq=\frac{c}{a}.\frac{1}{k^{2} }pq=
a
c
.
k
2
1
applying root on both sides
\sqrt{pq}=\sqrt{ \frac{c}{a}}.\frac{1}{k}
pq
=
a
c
.
k
1
----------------(2)
Dividing (1) by (2)
\frac{p+q}{\sqrt{pq} }=\frac{(-c/a)(1/k)}{\sqrt{\frac{c}{a} }(1/k) }
pq
p+q
=
a
c
(1/k)
(−c/a)(1/k)
\frac{p}{\sqrt{pq} }+\frac{q}{\sqrt{pq} }=-\sqrt{\frac{c}{a}}.\sqrt{\frac{c}{a}}/\sqrt{\frac{c}{a}}
pq
p
+
pq
q
=−
a
c
.
a
c
/
a
c
\frac{\sqrt{p}\sqrt{p} }{\sqrt{pq}}+\frac{\sqrt{q}\sqrt{q} }{\sqrt{pq}}=-\sqrt{\frac{c}{a} }
pq
p
p
+
pq
q
q
=−
a
c
\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}=-\sqrt{\frac{c}{a}}
q
p
+
p
q
=−
a
c
\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{c}{a}}=0
q
p
+
p
q
+
a
c
=0