If the equation x2-y2-10x-21 - 0 has real roots x - - and y - - then

The correct options are C 3 ≤ x ≤ 7 D 2 ≤ y ≤ 2 x^2 10x+21+y^2=0 x=10±√(100 4(21+y^2))2 =10±√(16 4y^2)2 =5±√(4 y^2) For real roo ...

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Byju's Answer

Standard XII

Mathematics

Relations between Roots and Coefficients : Higher Order Equations

If the equati...

Question

If the equation $x^{2} + y^{2} - 10 x + 21 = 0$ has real roots $x = α$ and $y = β$ then

3 \leq x \leq 7

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3 \leq y \leq 7

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- 2 \leq y \leq 2

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- 2 \leq x \leq 2

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Solution

The correct options are
C $3 \leq x \leq 7$
D $- 2 \leq y \leq 2$
$x^{2} - 10 x + 21 + y^{2} = 0$

$x = \frac{10 \pm \sqrt{100 - 4 (21 + y^{2})}}{2}$
$= \frac{10 \pm \sqrt{16 - 4 y^{2}}}{2}$
$= 5 \pm \sqrt{4 - y^{2}}$

For real roots

$4 - y^{2} \geq 0$
or
$y^{2} \leq 4$
or
$- 2 \leq y \leq 2$ ... $(i)$

Now

$x_{max} = (5 + \sqrt{4 - y^{2}})_{y = 0}$

$= 5 + 2$

$= 7$ .

And

$x_{min} = (5 - \sqrt{4 - y^{2}})_{y = 0}$

$= 5 - 2$

$= 3$ .

Hence
$3 \leq x \leq 7$

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If the equation x2+y2−10x+21=0 has real roots x=α and y=β then

The correct options are C 3≤x≤7 D −2≤y≤2x2−10x+21+y2=0x=10±√100−4(21+y2)2=10±√16−4y22=5±√4−y2For real roots4−y2≥0or y2≤4or −2≤y≤2 ...(i)Now xmax=(5+√4−y2)y=0=5+2=7.And xmin=(5−√4−y2)y=0=5−2=3.Hence 3≤x≤7

If the equation $x^{2} + y^{2} - 10 x + 21 = 0$ has real roots $x = α$ and $y = β$ then