Solve the inequality. Express your answer in interval form. $x ^ { 2 } - 4 x + 1 \leq 0$
A) $( 2,2 + \sqrt { 3 } ]$
B) $( - \infty , 2 - \sqrt { 3 } ] \cup [ 2 + \sqrt { 3 } , \infty )$
C) $[ 2 - \sqrt { 3 } , 2 + \sqrt { 3 } ]$
D) $( 0,2 - \sqrt { 3 } ] \cup [ 2 + \sqrt { 3 } , 0 )$
E) no solution

Question

Quizplus · Accepted Answer

We can solve the inequality by factoring: $(x-2\pm\sqrt{3})(x-2\mp\sqrt{3})\leq 0$. The roots are $2\pm\sqrt{3}$, so we test the intervals $(-\infty,2-\sqrt{3}]$, $(2-\sqrt{3},2+\sqrt{3}]$, and $[2+\sqrt{3},\infty)$ by plugging in test points. We can conclude that the solution is $[2-\sqrt{3},2+\sqrt{3}]$.

Solve the Inequality x2−4x+1≤0x ^ { 2 } - 4 x + 1 \leq 0x2−4x+1≤0

Solve the Inequality $x ^ { 2 } - 4 x + 1 \leq 0$