if x² - ax + b and x² - cx + d are divisible by (x - m), prove that m = (d-b)/(c-a)
Answers
Step-by-step explanation:
for this problem we will use the property of the sum and product of roots of a quadratic
that is
if
α
&
β
are the roots of
p
x
2
+
q
x
+
r
=
0
then
α
β
=
−
q
p
α
β
=
r
p
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x
2
+
a
x
+
b
=
0
−
−
−
(
1
)
x
2
+
c
x
+
d
=
0
−
−
−
(
2
)
let the common root be
α
for eqn
(
1
)
α
+
α
=
−
a
⇒
α
=
−
a
2
&
α
2
=
b
for the eqn
(
2
)
let the second root be
β
then
α
+
β
=
−
c
α
β
=
d
⇒
β
=
d
α
∴
α
+
d
α
=
−
c
α
2
+
d
=
α
(
−
c
)
b
+
d
=
(
−
a
2
)
(
−
c
)
∴
2
(
b
+
d
)
=
a
c
as reqd.
Answer:
Refer to the following solution:
Step-by-step explanation:
Let f(x) = x²-ax+b and p(x) = x²-cx+d
We have,
(x-m) is a factor of f(x). [since f(x) is divisible by (x-m)]
Therefore,by factor theorem
f(m)=0. [since, x-m= 0 or, x= m]
or, m²- am+ b= 0..............(1)
Again,
(x-m) is a factor of p(x). [since p(x) is divisible by (x-m)]
Therefore,by factor theorem
p(m)=0. [ since, x-m= 0 or, x= m]
or, m²-cm+d= 0 .................(2)
From (1) and (2) we get
m²-am+b= m²-cm+d
or, -am + b = -cm + d
or, cm - am = d - b
or, (c-a)m = (d-b)
or, m = (d-b)/(c-a). [Proved]