Let x , y , z be real variables satisfying the equations x + y + z = 6 and xy + | TestHub

Passage / Comprehension

$y = a x^{2} + b x + c = 0$ , for all $a, b, c \in R$ with $a \neq = 0$ , is a quadratic equation which has real roots if and only if $b^{2} - 4 a c \geq 0$ . If $F (x, y) = 0$ is a second-degree equation, then using the above fact, we can determine the range of $x$ and $y$ by treating it as a quadratic equation in $y$ or $x$ . Similarly, $a x^{2} + b x + c \geq 0$ for all $x \in R$ if $a > 0$ and $b^{2} - 4 a c \leq 0$ .

Let $x, y, z$ be real variables satisfying the equations $x + y + z = 6$ and $xy + yz + zx = 7$ , then the range of x is

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Answer:

Solution:

We have $x + y + z = 6$ .........(i)

$x y + yz + z x = 7 \dots\dots\dots (ii)$

From (i), $z = 6 - x - y$ & putting it in (ii),

we get $x y + y (6 - x - y) + x (6 - x - y) = 7$

or $y^{2} + (x - 6) y + (x^{2} - 6 x + 7) = 0$

Since y is real, $(x - 6)^{2} - 4 (x^{2} - 6 x + 7) \geq 0$

$\Rightarrow 3 x^{2} - 12 x - 8 \leq 0 \Rightarrow \frac{6 - 2 15}{3} \leq x \leq \frac{6 + 2 15}{3}$

Mathematics - Quadratic Equation Question with Solution | TestHub

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Doubts & Discussion

JEE Advanced 2028 (Class 11)