E-Storeno. of real solutions....u know?
We all find no. of real solutions for a 2 degree equation....
but wat if the ask u like a polynomial of degree 8 , they ask u to find the no. of real roots...+ive or -ive.... or complex roots....
dont worry...herez a sol. (some of u might know ) but rest.....look here
for eg. equation is
x4 - 3x + 1 = 0....
we can write it as
(+) x4 - (3x) + (1) = 0...
see the sign change.....
+ to - , then - to + , 2 changes na
so no. of +ive solutions = 2
now to -ive solutions
we gotf(x) = x4 + 3x + 1 = 0....
take f(-x) = x4 + 3x +1 = 0.....
no sign changes......so no -ive root
for complex roots , see the highest degree.....i.e 4 here
so complex roots = highest degree - (no. of + ive + no. of -ive roots)
= 4 - (2+0) = 2 complex roots
hope u got it....ok so now post the no. of +ive , -ive n complex roots =
x8 - 4x7 + 3x6 + x5 - x4 + x3 - 16x + 2 = 0
1 thing more....
if equation of type ax2 + b |x| + c = 0 , no. of real solutions is = 0
and when roots of an equation are equal in magnitude but of opp. sign then sum of roots is = to zero......
hope it helps....
cheers!!!!!!!!
ALL THE BEST :)
Comments (19)
thanx a lot for posting...
keep up the gr8 work...
i dun have tmh O:-)
read them else where....n its good for those who dun have TMH (like me)
and thanx for the comments all....
himanshu
well there
u left out one thing...........
to find no. of complex roots.........
we have to check if zero is also a root of the given equation........
and i have a doubt..... why cant the equation x4 - 3x + 1 have 4 complex roots and 0 positive roots.........
gives the MAXIMUM NO OF +ve/-ve ROOTS
so u CANNOT SAY that there are 2 +ve roots fr 2 conseq changes
in sign in f(x) or 2 -ve roots fr 2 conseq changes in sign in f(-x) but u can say that there WILL be MAXIMUM 2 +ve and 2-ve roots
to illustrate the above statement:
eg: f(x)=x^2+x+1 0 changes in sign so 0+ve roots
f(-x)=x^2-x+1 2 changes in sign=>2 ive roots WHICH IS WRONG
u all must be knowing that f(x) has 2 complex roots w and w^2
ONE CAN ONLY CORRECTLY SAY THAT THELL BE NO +VE ROOTS
IN THE EXAMPLE U HAVE ASKED THE GOIITIANS TO ANS
max no of +ve roots=6
max no of -ve roots=2
no of complex roots=0(sry i cant exactly say whether it wod be max or minm becoz it may vary frm q to q)
SO THE METHOD STATED BY U WILL BE USEFUL TO KNOW IF THERE ARE NO +VE OR NO -VE SOLN IN OBJ QUESTIONS bt it may lead to ERRONEOUS results IF USED OTHERWISE TO KNOW NO OF REAL SOLUTIONS.
SO as a true frnd i wod just say - Himanshu and all the other goiitians pls be very cautious of such shortcuts given in books without their limitations LAY MORE STRESS ON CONCEPTS , such things are not evn mentioned in jee syllabus. NO SUCH QUESTION WILL BE ASKED IN HE EXAM which cannot be solvd using the basic concepts and confining to the constraints of the jee syllabus.
wen solving we get x=-1 n 1 ....the reason for ax^2 +modx*b+c has no real solutions wen all a b c are positive is tht...wen such a quadratic equation with all postive coefficents hav negative roots.. in our question d variable is mod x ....mod x cant b negative...tht y no solutions....nice article but i think u explained everything niclyyy...u shld hav added tht all a b c r positive.....oth wise ppl may think its a general rule....nice work anyway...
hats off.....and sorry :(
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