Find the total number of real roots of below-given equation
(?+?)^?+ (?+?)^?= ??
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3
4
No solution
2 ✔✔
4
X= -3or -5 so answer is 2
4 real roots
Using Descartes theorem
-3, -5, -7, -9
3 -'ve, 1+ve= total 4
Sir , no values simultaneously satisfy both parts of the equation.Hence ,the answer is ''0''.
-3,-5
4
sir radius one kaise mani
3
4
4
2 real roots
2 roots
No real roots possible
no real roots possible
-12+root124/2 and-12-root124/2
Two real roots [-6+√(✓24)-3]. And[-6√(✓24)-3]
Awesome
-7,-3,-9,-5 are the roots since equation is of degree 4
5+2√35, 5-2√35, 7+2√35, 7-2√35 are 4 roots
4 real roots (All are negative)
2
x=-3 ,-5, -7 ,-9
-5 ,-7
(x+5)^4+(x+7)^4=2^4+2^4
X+5=+/-2, x+7=+/-2
X=-2+-5=-7,x=2-5=-3,x=-2-7=-9,x=2-7=-5
X=-3,-5,-7,-9
Expand and factorize to get:
(x²+12x+39)²=24
Now, solve to get
X²+12x+36=2rt6-3, x²+12x+39=-2rt6
Solve to get:
X=-6±rt(2rt6-3)
So total number of real roots are 2.
-3,-7,-5,-9
Super sir aap online course available hai kya
0
2 real roots
It must be 2
On breaking 32 into two parts i.e 2^4+2^4
we get x=-3 and -5
SIR I WANT ENQUIRY OF OFF LINE CLASS
NO REAL ROOT POSIBLE
There are no real roots, hence answer is 0(zero)
There are 4 real roots one positive and three negetive roots by descarte's rule of signs…
real roots will be 2 first root is between -4 and -5 second root is between -7 and -8
root can be find by newton rapshon method