solve the following equation: (2x+3)(2x+5)(x-1)(x-2)=30
solve please this question
Answers
Answer:
Let us evaluate,
\displaystyle\sf{\longrightarrow I=\int\dfrac{1}{(2\sin x+\sec x)^4}\ dx}⟶I=∫
(2sinx+secx)
4
1
dx
\displaystyle\sf{\longrightarrow I=\int\dfrac{1}{\left(2\sin x+\dfrac{1}{\cos x}\right)^4}\ dx}⟶I=∫
(2sinx+
cosx
1
)
4
1
dx
\displaystyle\sf{\longrightarrow I=\int\dfrac{\cos^4x}{\left(1+\sin(2x)\right)^4}\ dx}⟶I=∫
(1+sin(2x))
4
cos
4
x
dx
\displaystyle\sf{\longrightarrow I=\int\dfrac{\cos^4x}{\left(\left(\sin x+\cos x\right)^2\right)^4}\ dx}⟶I=∫
((sinx+cosx)
2
)
4
cos
4
x
dx
\displaystyle\sf{\longrightarrow I=\int\dfrac{\cos^4x}{\left(\sin x+\cos x\right)^8}\ dx}⟶I=∫
(sinx+cosx)
8
cos
4
x
dx
\displaystyle\sf{\longrightarrow I=\int\dfrac{\dfrac{\cos^4x}{\cos^8x}}{\left(\dfrac{\sin x+\cos x}{\cos x}\right)^8}\ dx}⟶I=∫
(
cosx
sinx+cosx
)
8
cos
8
x
cos
4
x
dx
\displaystyle\sf{\longrightarrow I=\int\dfrac{\sec^4x}{\left(1+\tan x\right)^8}\ dx}⟶I=∫
(1+tanx)
8
sec
4
x
dx
\displaystyle\sf{\longrightarrow I=\int\dfrac{\sec^2x\cdot\sec^2x}{\left(1+\tan x\right)^8}\ dx}⟶I=∫
(1+tanx)
8
sec
2
x⋅sec
2
x
dx
\displaystyle\sf{\longrightarrow I=\int\dfrac{1+\tan^2x}{\left(1+\tan x\right)^8}\cdot\sec^2x\ dx}⟶I=∫
(1+tanx)
8
1+tan
2
x
⋅sec
2
x dx
\displaystyle\sf{\longrightarrow I=\int\dfrac{(1+\tan x)^2-2\tan x}{\left(1+\tan x\right)^8}\cdot\sec^2x\ dx}⟶I=∫
(1+tanx)
8
(1+tanx)
2
−2tanx
⋅sec
2
x dx
Substitute \sf{u=1+\tan x.}u=1+tanx.
\displaystyle\sf{\longrightarrow I=\int\dfrac{u^2-2(u-1)}{u^8}\ du}⟶I=∫
u
8
u
2
−2(u−1)
du
\displaystyle\sf{\longrightarrow I=\int\dfrac{u^2-2u+2}{u^8}\ du}⟶I=∫
u
8
u
2
−2u+2
du
\displaystyle\sf{\longrightarrow I=\int\dfrac{1}{u^6}\ du-2\int\dfrac{1}{u^7}\ du+2\int\dfrac{1}{u^8}\ du}⟶I=∫
u
6
1
du−2∫
u
7
1
du+2∫
u
8
1
du
\displaystyle\sf{\longrightarrow I=\dfrac{u^{-5}}{-5}-2\cdot\dfrac{u^{-6}}{-6}+2\cdot\dfrac{u^{-7}}{-7}+k}⟶I=
−5
u
−5
−2⋅
−6
u
−6
+2⋅
−7
u
−7
+k
Undoing the substitution,
\displaystyle\sf{\longrightarrow I=-\dfrac{1}{5}(1+\tan x)^{-5}+\dfrac{1}{3}(1+\tan x)^{-6}-\dfrac{2}{7}(1+\tan x)^{-7}+k\quad\quad\dots(1)}⟶I=−
5
1
(1+tanx)
−5
+
3
1
(1+tanx)
−6
−
7
2
(1+tanx)
−7
+k…(1)
But in the question it's given,
\displaystyle\sf{\longrightarrow I=A(1+\tan x)^{-5}+B(1+\tan x)^{-6}+C(1+\tan x)^{-7}+k\quad\quad\dots(2)}⟶I=A(1+tanx)
−5
+B(1+tanx)
−6
+C(1+tanx)
−7
+k…(2)
Comparing (1) and (2) we get,
\sf{A=-\dfrac{1}{5}}A=−
5
1
\sf{B=\dfrac{1}{3}}B=
3
1
\sf{C=-\dfrac{2}{7}}C=−
7
2
Therefore,
\sf{\longrightarrow A+B+C=-\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{2}{7}}⟶A+B+C=−
5
1
+
3
1
−
7
2
\sf{\longrightarrow\underline{\underline{A+B+C=-\dfrac{16}{105}}}}⟶
A+B+C=−
105
16
Hence (D) is the answer
Step-by-step explanation:
( 4x² +10x+6x+15) (x²-2x-x+2)=30
4x²+16x+15) ( x²-3x+2)=30
4x⁴-12x³+8x²+16x³-48x²+32x+15x²-45x+30=30
x²(4x²-25)+4x³-13x+30=30
x²(4x²-25)+4x³-13x=0
4x²-25+4x³-13x=0
4x²+4x³-13x=25
x(4x+4x²-13)-25=0
4x+4x²-38=0
4x(1+x)-38=0
1+x-38=0
x-37=0
x=37