Question

The set of values of c so that the equation y = |x| + c and $x^2+y^2-8\left|x\right|-9=0$ have no solution is

• A. $(-\infty ,-3)\cup (3,\infty )$
• B. (-3, 3)
• C. $(-\infty ,-5\sqrt{2})\cup (5\sqrt{2},\infty )$
• D. $(-\infty ,-4-5\sqrt{2})\cup (5\sqrt{2}-4,\infty )$

Answer 1

The correct option is D $(-\infty ,-4-5\sqrt{2})\cup (5\sqrt{2}-4,\infty )$

Given equation

$y=\left|x\right|+c$---(1)

$x^2+y^2-8\left|x\right|-9=0$

From equation (1) and (2)

$x^2+(\left|x\right|+c{)}^2-8\left|x\right|-9=0$

when $x>0$

$x^2+(x+c{)}^2-8x-9=0$

$x^2+x^2+c^2+2cx-8x-9=0$

$2x^2+x(2c-8)+c^2-9=0$

For no solution

$D<0$

$(2c-8{)}^2-4\times 2(c^2-9)<0$

$4c^2+64-32c-8c^2+72<0$

$-4c^2-32c+136<0$

$c^2+8c-34>0$

$c=\frac{-8\pm \sqrt{64+136}}{2}$

$c=\frac{-8\pm \sqrt{200}}{2}$

$c=-4\pm 5\sqrt{2}$

C has root $c=-4\pm 5\sqrt{2}$

Hence for no solution c has all value excluding it's roots

$cϵ(-\infty ,-4-5\sqrt{2})\cup (5\sqrt{2}-4,\infty )$