Quadratic Trouble #1
1 2 x 4 − 5 6 x 3 + 8 9 x 2 − 5 6 x + 1 2 = 0
Find the sum of real roots x that satisfy the equation above. Give your answer to 2 decimal places.
Try my other algebra problems here .
The answer is 4.67.
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6 solutions
Moderator note:
Wonderful, this is the solution I'm looking for. You just need to apply Rational Root Theorem (RRT) "half way" and show that the remaining roots are real as well. Then the answer is immediate by Vieta's formula.
Here's a few tricks to reduce your work: Let f ( x ) denote the given polynomial, and notice that x 4 f ( x 1 ) = 0 is equivalent to f ( x ) = 0 , then if x = p is a root, x = p 1 is also root (Do you see why?). By RRT, we get x = 2 as a possible solution, then x = 2 1 is also a possible solution. Can you see how to spot polynomials like this?
@Challengemaster , x = p . x = p 1 are the roots of polynomial , of degree n , if and only if the coefficients of x a and x n − a , where a is a non-negative integer, are equal for all a = 0 , 1 , 2 , . . . . , n .
VIGGY,tussi great ho..
56 / 12 = 4.67 vieta
fab solution, just noted that in my math book.
Firstly, I plugged a few numbers like (-1, 1, -2 and 2) into f(x). Only f(2)=0.
If f(2)=0 ... then (x-2) is a factor.
I divided the equation by (x-2), by polynomial division, to give:
1 2 x 3 − 3 2 2 + 2 5 x − 6
I then plugged a few numbers into the above equation until f(x)=0. I found that f(1.5)=0. Therefore, (2x-3) is a factor.
I divided the above equation by (2x-3) to give:
6 x 2 − 7 x + 2
By inspection, I "saw" that this could be factorised as (3x-2)(2x-1). x=2/3, 1/2
Therefore, our roots are 2, 1.5, 2/3 and 1/2. Sum=14/3 or 4.66
Moderator note:
Right, this is the most standard approach.
Same way bro..But do we have any simpler way of solving higher degree polynomial.
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In this case we have
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I tried but.....would you please help me@Nihar Mahajan@Rajdeep Dhingra
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@Sachin Mittal – I have given the solution as a comment to Sir(Nihar sir).
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@Rajdeep Dhingra – I am not capable enough to accept you summon me as 'Sir'.You can just call me "Nihar".
Challenge student note : Just observe the coefficients of the polynomial. Do you seek any symmetry? So can you devise a simpler method to solve this problem ?
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Yes sir, I can
The polynomial is
1
2
x
4
−
5
6
x
3
+
8
9
x
2
−
5
6
x
+
1
2
=
0
Dividing both sides by
x
2
we get
1
2
x
2
−
5
6
x
+
8
9
−
5
6
x
1
+
1
2
x
2
1
=
0
1
2
(
(
x
+
x
1
)
2
−
2
)
−
5
6
(
x
+
x
1
)
+
8
9
=
0
L
e
t
x
+
x
1
=
y
1
2
y
2
−
5
6
y
+
6
5
=
0
Applying quadractic formula we can solve for the roots.
We can then equate it to
x
+
x
1
And again solve a quadratic. Cheers We got are roots.
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Why did you let y = x + x 1 ?
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@James Cochrane – To make it a quadratic . Sir as we have the formula for quadratic we can evaluate that but not a bi-quadratic.
Nice solution. Alludes to transformation.
How did I become a Sir from student ? O.o
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@Nihar Mahajan – Sir U are great. I am not capable enough to call U by name or as a student.
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@Rajdeep Dhingra – Well then I am also not capable enough to accept you summon me as 'Sir'.You can just call me "Nihar".
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@Nihar Mahajan – No Sir U I will always call U a sir.
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@Rajdeep Dhingra – Okay , then I will call you Raju. (Remember it is undiplomatic :P)
Just use VIETA'S THEOREM -B/A = 56/12 = 4.67
Actually you were lucky this worked and there were no complex roots with real parts.
Use Vieta's Relation
Divide the coefficient of second term to the leading coefficient. ( 1 2 5 6 ) = 4 . 6 6 6 6 6 7
for sum of roots:
sum of roots of a 4th degree polynomial eq.=-(co-efficient of x3 term) /co-efficient of x4 term
so,answer is-(-56)/12=4.67
x3 is x cube and x4 is x to the power of 4
Firstly divide the whole equation by x^2... Therefore we get a quadratic equation in x...now to solve the equation firstly we verify the discriminant...now see the tricky part....the sign on 59 is negative and it forms a part of c in b^2-4ac..and thus the negative sign of -4 is cancelled out by -59...and thus all roots are real and thus we proceed further and use the vieta formulae for the new quadratic equation formed after dividing by x^2 and sum of roots is -b/a and thus it follows 56/12=4.66
Moderator note:
You're wrong.
Firstly divide the whole equation by x^2... Therefore we get a quadratic equation in x
This is not true, because the equation 1 2 x 2 − 5 6 x + 8 9 − x 5 6 + x 2 1 2 = 0 is not a quadratic equation because it is not in the form of A x 2 + B x + C = 0 .
Okkk....but I had no knowledge of RRT so I did this...is there any method other than RRT method... @CHALLENGE MASTER
but the equation obtained by dividing the initial equation by x^2 can be reduced to am quadratic in terms of(x+1/x) @CHALLENGE MASTER
Here, we will use rational root test . The Rational root theorem tells you that if the polynomial has a rational zero then it must be a fraction q p , where p is factor of the trailing constant and q is the factor of leading coefficient. So, the Rational Roots Tests yields the possible solutions { ± q p } p , q = 1 , 2 , 3 , 4 , 6 , 1 2 . If we plug these values into the polynomial P ( x ) , we get P ( 2 1 ) = 0 . Also, P ( 2 ) = 0 . Hence, P ( x ) = ( 2 x − 1 ) ( x − 2 ) ( 6 x 2 − 1 3 x + 6 ) . The discriminant of the quadratic in P ( x ) is 1 3 2 − 4 ( 6 ) ( 6 ) = 2 5 = p o s i t i v e . Now, we know that all roots of the polynomial are real, so by using VIETA's formula , sum of real roots of the polynomial is − a b = − 1 2 ( − 5 6 ) = 4 . 6 6 7