Question:

If the equation $x^2 + y^2 -10x + 21 = 0$ has real roots $x = a$ and $y=\beta$ then

Updated On: Jun 18, 2022
  • $3\le x\le7$
  • $3\le y\le7$
  • $-2 \le y \le2$
  • $-2\le x\le2$
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The Correct Option is C

Solution and Explanation

$x^{2}-10x+\left(y^{2}+21\right)=0$
for real roots of $x, D \ge0$
$100-4\left(y^{2}+21\right)\ge0$
$\Rightarrow y^{2}\le4$
$\Rightarrow -2\le y\le2 \left(C\right)$
also, $y^{2}=-x^{2}+10x-21$
for real roots of $y,$
$-x^{2}+10x-21 \ge0$
$\Rightarrow \left(x-7\right)\left(x-3\right)\le0$
$3\le x\le7 \left(A\right)$
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Concepts Used:

Complex Numbers and Quadratic Equations

Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.

Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.