Find the values of b such that the system $$\left\{ \begin{aligned}
x ^ { 2 } + y ^ { 2 } & = 9 \
y & = x + b
\end{aligned} \right.$$ has no solution.
A) $( - \infty , - 3 ) \cup ( 3 , \infty )$
B) $( - \infty , \infty )$
C) $( - \infty , - 3 \sqrt { 2 } ) \cup ( 3 \sqrt { 2 } , \infty )$
D) $( - 3 \sqrt { 2 } , 3 \sqrt { 2 } )$
E) no solution

Question

Quizplus · Accepted Answer

Substitute the second equation into the first equation:$$x^2 + (x + b)^2 = 9$$$$x^2 + x^2 + 2bx + b^2 = 9$$$$2x^2 + 2bx + b^2 - 9 = 0$$The discriminant of this quadratic equation is:$$b^2 - 4(2)(b^2 - 9) = b^2 - 8(b^2 - 9) = -7b^2 + 72$$For the system to have no solution, the discriminant must be negative. Therefore, we must solve the inequality:$$-7b^2 + 72 < 0$$$$-7b^2 < -72$$$$b^2 > 12$$$$b < -3\sqrt{2} ext{ or } b > 3\sqrt{2}$$Therefore, the values of b such that the system has no solution are:$$(-\infty, -3\sqrt{2}) \cup (3\sqrt{2}, \infty)$$

Find the Values of B Such That the System {x2+y2=9y=x+b\left\{ \begin{aligned}x ^ { 2 } + y ^ { 2 } & = 9 \\y & = x + b\end{aligned} \right.{x2+y2y​=9=x+b​

Find the Values of B Such That the System $\left\{ \begin{aligned}x ^ { 2 } + y ^ { 2 } & = 9 \\y & = x + b\end{aligned} \right.$