one roof of the quadratic equation x square + px + 3 = 0 is 1 other root is.
Answers
Step-by-step explanation:
\large\underline{\sf{Solution-}}
Solution−
Given parametric functions are
\begin{gathered}\rm \: x = \sqrt{ {a}^{ {cos}^{ - 1}t } } \: - - - (1) \\ \end{gathered}
x=
a
cos
−1
t
−−−(1)
and
\begin{gathered}\rm \: y = \sqrt{ {a}^{ {sin}^{ - 1}t } } \: - - - (2) \\ \end{gathered}
y=
a
sin
−1
t
−−−(2)
On multiply equation (1) and (2), we get
\begin{gathered}\rm \: xy = \sqrt{ {a}^{ {cos}^{ - 1}t } } \times \sqrt{ {a}^{ {sin}^{ - 1}t } } \\ \end{gathered}
xy=
a
cos
−1
t
×
a
sin
−1
t
We know,
\begin{gathered}\boxed{\sf{ \: {x}^{m} \times {y}^{m} \: = \: {(xy)}^{m} \: }} \\ \end{gathered}
x
m
×y
m
=(xy)
m
So, using this identity, we get
\begin{gathered}\rm \: xy = \sqrt{ {a}^{ {cos}^{ - 1}t } \times {a}^{ {sin}^{ - 1} t} } \\ \end{gathered}
xy=
a
cos
−1
t
×a
sin
−1
t
We know,
\begin{gathered}\boxed{\sf{ \: {x}^{m} \times {x}^{n} \: = \: {x}^{m + n} \: }} \\ \end{gathered}
x
m
×x
n
=x
m+n
So, using this identity, we get
\begin{gathered}\rm \: xy = \sqrt{ {a}^{ {cos}^{ - 1}t + {sin}^{ - 1} t}} \\ \end{gathered}
xy=
a
cos
−1
t+sin
−1
t
We know,
\begin{gathered}\boxed{\sf{ \: {cos}^{ - 1}x + {sin}^{ - 1}x = \frac{\pi}{2} \: }} \\ \end{gathered}
cos
−1
x+sin
−1
x=
2
π
So, using this identity, we get
\rm \: xy \: = \: \sqrt{ {\bigg(a\bigg) }^{\dfrac{\pi}{2}} }xy=
(a)
2
π
On dividing both sides w. r. t. x, we get
\begin{gathered}\rm \: \dfrac{d}{dx}(xy) \: = \: \dfrac{d}{dx}\sqrt{ {\bigg(a\bigg) }^{\dfrac{\pi}{2}} } \\ \end{gathered}
dx
d
(xy)=
dx
d
(a)
2
π
\begin{gathered}\rm \: x\dfrac{d}{dx}y + y\dfrac{d}{dx}x = 0 \\ \end{gathered}
x
dx
d
y+y
dx
d
x=0
\begin{gathered}\rm \: x\dfrac{dy}{dx} + y \times 1 = 0 \\ \end{gathered}
x
dx
dy
+y×1=0
\begin{gathered}\rm \: x\dfrac{dy}{dx} + y = 0 \\ \end{gathered}
x
dx
dy
+y=0
\begin{gathered}\rm \: x\dfrac{dy}{dx}= - y\\ \end{gathered}
x
dx
dy
=−y
\begin{gathered}\rm\implies \:\dfrac{dy}{dx} \: = \: - \: \dfrac{y}{x} \\ \end{gathered}
⟹
dx
dy
=−
x
y
Hence, Proved
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Additional Information
\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}\end{gathered}
f(x)
k
sinx
cosx
tanx
cotx
secx
cosecx
x
logx
e
x
dx
d
f(x)
0
cosx
−sinx
sec
2
x
−cosec
2
x
secxtanx
−cosecxcotx
2
x
1
x
1
e
x
Answer:
P + X + 3 = 0 1 2 3 0 1 1 3 + 3 y + 2 = 0 1 root is multiple + 4 = 0 1 roots multiple 5 = 0 root 1 multiple + 3 = 0 and multiple 1