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If x, y and z are real numbers such that x + y + z = 5 and xy + yz + zx = 3, what is the largest value that x can have?
- a)5/3
- b)13/3
- c)√19
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
If x, y and z are real numbers such that x + y + z = 5 and xy + yz + z...
The given equations are x + y + z = 5 — (1) , xy + yz + zx = 3 — (2)
xy + yz + zx = 3
x(y + z) + yz = 3
⇒ x ( 5 - x ) +y ( 5 – x – y) = 3
⇒ -y2 - y(5 - x) - x2 + 5x = 3
⇒ y2 + y(x - 5) + (x2 - 5x + 3) = 0
The above equation should have real roots for y, => Determinant >= 0
⇒ y2 + y(x - 5) + (x2 - 5x + 3) = 0
The above equation should have real roots for y, => Determinant >= 0
⇒ b2 - 4ac0
⇒ (x - 5)2 - 4(x2 - 5x + 3) ≥ 0
⇒ 3x2 - 10x - 13 ≤ 0
⇒ -1 ≤ x ≤ 13/3
Hence maximum value x can take is 13/3, and the corresponding values for y,z are 1/3, 1/3
Most Upvoted Answer
If x, y and z are real numbers such that x + y + z = 5 and xy + yz + z...
Step :-1)
From x + y + z = 5, we get y + z = 5-x.
From xy + yz + zx = 3, we get yz + x(y + z) = 3.
Substitute y + z: yz + x(5 – x) = 3.
So, yz = 3 - 5x + x^2
Step:-2)
Product of the roots = yz = 3-5x+x^2
Sum of the roots = y+z = 5-x
y and z are roots of the quadratic equation t^2 - (y + z)t + yz = 0.
Substitute the expressions from Step 1: t^2 - (5-x)t + (x2 – 5x + 3) = 0.
Step:-3)
Since y and z are real numbers, the discriminant of the quadratic equation must be non-negative.
D= b^2 - 4ac≥0
D = (-(5 – x))^2 - 4(1)(x^2 – 5x + 3) ≥ -0.
-3x^2+10x+13 ≥ 0.
Multiply by -1 and reverse the inequality: 3x^2 - 10x -13 ≤ 0
Step :-4)
Roots of the equation
3x^2 -10x -13≤0
x=-1 & x= 13/3
-1≤x≤13/3
From x + y + z = 5, we get y + z = 5-x.
From xy + yz + zx = 3, we get yz + x(y + z) = 3.
Substitute y + z: yz + x(5 – x) = 3.
So, yz = 3 - 5x + x^2
Step:-2)
Product of the roots = yz = 3-5x+x^2
Sum of the roots = y+z = 5-x
y and z are roots of the quadratic equation t^2 - (y + z)t + yz = 0.
Substitute the expressions from Step 1: t^2 - (5-x)t + (x2 – 5x + 3) = 0.
Step:-3)
Since y and z are real numbers, the discriminant of the quadratic equation must be non-negative.
D= b^2 - 4ac≥0
D = (-(5 – x))^2 - 4(1)(x^2 – 5x + 3) ≥ -0.
-3x^2+10x+13 ≥ 0.
Multiply by -1 and reverse the inequality: 3x^2 - 10x -13 ≤ 0
Step :-4)
Roots of the equation
3x^2 -10x -13≤0
x=-1 & x= 13/3
-1≤x≤13/3
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Community Answer
If x, y and z are real numbers such that x + y + z = 5 and xy + yz + z...
Let's call the three numbers x, y, and z. We are given that x+y+z=5 and xy+yz+zx=3.
We can rewrite the first equation as x+y+z=5 as xy+yz+zx+2xy+2yz+2zx=3+2xy+2yz+2zx.
Adding the two equations together, we have 3+2xy+2yz+2zx=3+3+2xy+2yz+2zx.
Simplifying, we have 6=6+2xy+2yz+2zx.
Subtracting 6 from both sides, we have 0=2xy+2yz+2zx.
Dividing both sides by 2, we have 0=xy+yz+zx.
We can rewrite this equation as x(y+z)+yz=0.
Since x, y, and z are real numbers, we know that y+z and yz must have opposite signs (if they had the same sign, then their sum would not be zero).
This means that either y and z are both negative, or one of them is negative and the other is positive.
If both y and z are negative, then y+z is negative. Since the sum of three real numbers is 5, this would mean that x is positive.
However, if one of y or z is negative and the other is positive, then their sum y+z is positive. Since the sum of three real numbers is 5, this would mean that x is negative.
Therefore, x cannot be positive.
The largest value that x can have is 0.
Therefore, the answer is c) 0.
We can rewrite the first equation as x+y+z=5 as xy+yz+zx+2xy+2yz+2zx=3+2xy+2yz+2zx.
Adding the two equations together, we have 3+2xy+2yz+2zx=3+3+2xy+2yz+2zx.
Simplifying, we have 6=6+2xy+2yz+2zx.
Subtracting 6 from both sides, we have 0=2xy+2yz+2zx.
Dividing both sides by 2, we have 0=xy+yz+zx.
We can rewrite this equation as x(y+z)+yz=0.
Since x, y, and z are real numbers, we know that y+z and yz must have opposite signs (if they had the same sign, then their sum would not be zero).
This means that either y and z are both negative, or one of them is negative and the other is positive.
If both y and z are negative, then y+z is negative. Since the sum of three real numbers is 5, this would mean that x is positive.
However, if one of y or z is negative and the other is positive, then their sum y+z is positive. Since the sum of three real numbers is 5, this would mean that x is negative.
Therefore, x cannot be positive.
The largest value that x can have is 0.
Therefore, the answer is c) 0.
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