If the ratio of two roots of a quadratic equation ax²+bx+c=0[a≠0] is 1:r , then show that (r+1)²/r=b²/ac
Answers
+bx+c=0
roots are
2a
−b+
b
2
−4ac
α=
2a
−b+
b
2
4ac
β=
2a
−b−
b
2
−4ax
r=
β
α
=(
−b+
b
2
−4ac
−b+
b
2
4ac
)
=
−(b+
b
2
−4ac
)
b−
b
2
−4ac
×
b−
b
2
−4ac
b−
b
2
−4ac
=
b−b
2
+4ac
(b−
b
2
−4ac
)
2
=
4ac
b
2
+b
2
−4ac−2b
b
2
−4ac
=
4ac
2b
2
−2b
b
2
−4ac
−4ac
Step-by-step explanation:
Given :-
The ratio of two roots of a quadratic equation ax²+bx+c=0[a≠0] is 1:r .
To find :-
Show that (r+1)²/r=b²/ac ?
Solution :-
Given that
The Quadratic equation is ax²+bx+c = 0
a ≠ 0
Let A ,B are the two roots of the equation
Then
Sum of the zeroes = -b/a
=> A+B = -b/a --------(1)
Product of the zeroes = c/a
=> AB = c/a -----------(2)
Given that
The ratio of the roots = r:1
=> A:B = r:1
=> A/B = r/1
=> A/B = r
=> A = r× B ---------(3) or
=> B = A/r -----------(4)
On Substituting the value of A in (2)
=> r×B×B = c/a
=> r× B² = c/a
=> B² = c/ar --------(5)
On Substituting the value of B in (2)
=> A×A/r = c/a
=> A²/r = c/a
=> A² = cr/a ----------(6)
On squaring both sides of the equation (1)
=> (A+B)² = (-b/a)²
=> A²+B²+2AB = b²/a²
On Substituting values of A² ,B² then
=> (cr/a)+(c/ar)+2(c/a) = b²/a²
=> (cr/a)+(2c/a)+(c/ar) = b²/a²
=> (c/a)[r+2+(1/r)] = b²/a²
=>r+2+(1/r) = (b²/a²)×(a/c)
=>r+2+(1/r) = (ab²)/(a²c)
=>r+2+(1/r) = b²/ac
=> (r²+2r+1)/r = b²/ac
=> (r+1)²/r = b²/ac
Hence, Proved.
Answer:-
If the ratio of two roots of a quadratic equation ax²+bx+c=0[a≠0] is 1:r , then (r+1)²/r=b²/ac.
Used formulae:-
- Sum of the zeores = -b/a
- Product of the zeores = c/a
- (a+b)² = a²+2ab+b²