if a and b are the zeros of the polynomial px²+qx+r then find the values of 1/x + 1/b
Answers
Answer:
The value of (p+q+r)^2=q^2-4pr(p+q+r)
2
=q
2
−4pr
Step-by-step explanation:
The given quadratic poynomial:
px^2+qx+rpx
2
+qx+r
To find, the value of (p+q+r)^2=?(p+q+r)
2
=?
The sum of the roots,
\dfrac{B}{B-1}+\dfrac{B+1}{B}=-\dfrac{q}{p}
B−1
B
+
B
B+1
=−
p
q
The product of the roots,
\dfrac{B}{B-1}.\dfrac{B+1}{B}=\dfrac{r}{p}
B−1
B
.
B
B+1
=
p
r
∴ (\dfrac{B}{B-1}-\dfrac{B+1}{B})^2=(\dfrac{B}{B-1}+\dfrac{B+1}{B})^2-4(\dfrac{B}{B-1}.\dfrac{B+1}{B})(
B−1
B
−
B
B+1
)
2
=(
B−1
B
+
B
B+1
)
2
−4(
B−1
B
.
B
B+1
)
⇒ (\dfrac{B}{B-1}-\dfrac{B+1}{B})^2=(-\dfrac{q}{p} )^2-4(\dfrac{r}{p})(
B−1
B
−
B
B+1
)
2
=(−
p
q
)
2
−4(
p
r
)
⇒ (\dfrac{B}{B-1}-\dfrac{B+1}{B})^2=\dfrac{q^2-4pr}{p^2}(
B−1
B
−
B
B+1
)
2
=
p
2
q
2
−4pr
⇒ (p+q+r)^2=q^2-4pr(p+q+r)
2
=q
2
−4pr
Hence, the value of (p+q+r)^2=q^2-4pr(p+q+r)
2
=q
2
−4pr