Isolating A Single Variable!
⎩ ⎨ ⎧ x + y + z x y + y z + x z = 1 5 = 7 2
Let x , y , and z be real numbers satisfying the system of equations above.
Find the possible range of x .
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3 solutions
Used your method to solve! @Calvin Lin
I didn't understand how u took it as a quadratic??
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See the solution...the 2 n d line consists of the required quadratic which when factorized, yields the answer.
I've edited the problem for clarity, making a distinction between the variables and the solutions.
I am not able to view the right part of the solution
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I don't understand bro... What is 'right part of the solution'?
1) Right= Right side of the solution on your screen; or
2)Right=The correct method of the solution
Relevant wiki: Cubic Discriminant
Let f ( w ) = w 3 + A w 2 + B w + C be a cubic polynomial with roots x , y , z for constants A , B , C independent of its roots.
Then by Vieta's formula , A = − 1 5 , B = 7 2 . We can simplify f ( w ) to w 3 − 1 5 w 2 + 7 2 w + C .
Since f ( w ) has all real roots, then its discriminant is non-negative, that is
Δ 3 = b 2 c 2 − 4 a c 3 − 4 b 3 d − 2 7 a 2 d 2 + 1 8 a b c d ≥ 0 ,
where a = 1 , b = − 1 5 , c = 7 2 , d = C . Upon substitution, we get
− 2 7 ( C 2 + 2 2 0 C + 1 2 0 9 6 ) ≥ 0 ⇒ − 1 1 2 ≤ C ≤ − 1 0 8 .
When x is at the boundary points, then C = − 1 1 2 or − 1 0 8 .
If
C
=
−
1
1
2
, then factoring
f
(
w
)
gives
f
(
w
)
=
(
w
−
7
)
(
w
−
4
)
2
.
If
C
=
−
1
0
8
, then factoring
f
(
w
)
gives
f
(
w
)
=
(
w
−
6
)
2
(
w
−
3
)
.
Hence, the minimum root of either of these f ( w ) 's is at w = 3 , and the maximum root of either of these f ( w ) 's is at w = 7 .
Our answer is [ 3 , 7 ] .
Bonus : Did you notice that at both the boundary points, f ( w ) has a double root. Coincidence?
Great approach using the Cubic discriminant! I love this approach.
From the 2 eqn we can say... x² + y ² + z² = 81...and... x + y + z = 15.. O M G! Wait... Now it seems it's solvable using vectors... (x, y, z) is the vector... It's magnitude is 9 And it's dot with (1,1,1) gives 15. its simply a vector making a fixed angle with (1,1,1)....now find the highest and the least possible values of x - component.....which are 7 and 3 respectively.🙂simple.
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Can you write that up as a separate solution? It is not immediately obvious to me how you reach the final conclusion since there is a cone of vectors that make a fixed angle, and you want to intersect that with the sphere of magnitude 9.
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In the set of those cone of vectors ...there exists 2 unique vectors.... One with highest x-component and the other with least.
The highest and least x-components on solving comes out to be 7 and 3.
One can use few of the followings to find the 2 vectors (or their x components) ....
-
The vector (a) with highest x component will make largest angle with x axis.
-
The vector(b) with least x component will make least angle.
3.vectors (a) , (b) ,(1,1,1) and x axis will be coplanar.
From the 2 eqn we can say...
x² + y ² + z² = 81...and...
x + y + z = 15..
Now,
2(81 - x²) = 2(y² + z²) _> (y + z)² = (15 - x)²
Which implies that..... (x-3)(x-7) _< 0 Therefore... x lies in (3, 7) including both.
@Praveen Kumar I've converted it into a solution. Can you edit?
It's easy when you think about RMS >= AM
Suppose that X , Y , Z are real number solutions to the above system of equations.
Substitute Y = 1 5 − X − Z into the second equation. We get
X ( 1 5 − X − Z ) + ( 1 5 − X − Z ) Z + X Z = 7 2 ⇒ − Z 2 + ( 1 5 − X ) Z + ( − X 2 + 1 5 X − 7 2 ) = 0
Considering − z 2 + ( 1 5 − X ) z + ( − X 2 + 1 5 X − 7 2 ) = 0 as a quadratic in z . Since Z is a real solution, the quadratic discriminant is ≥ 0 , which implies
( 1 5 − X ) 2 − 4 ( − 1 ) ( − X 2 + 1 5 X − 7 2 ) ≥ 0
Expanding this, we obtain 3 ( X − 7 ) ( X − 3 ) ≤ 0 , and thus the solution set is 3 ≤ X ≤ 7 .
Up to this point, we have only shown that 3 ≤ X ≤ 7 is a necessary condition. We still have to show that it is sufficient. Conversely, given any X ∈ [ 3 , 7 ] , we can find a real Z that is a solution to − z 2 + ( 1 5 − X ) z + ( − X 2 + 1 5 X − 7 2 ) = 0 , and then set Y = 1 5 − X − Z . Then, it is clear that X , Y , Z satisfies the conditions in the problem.