Question

solve the inequality. write the solution set in interval notation.
$x^2 \leq 25$
what is the solution to the inequality? select the correct choice below and, if necessary, fill in the answer box to complete your choice.
\\(\bigcirc\\) a. the solution set is \\(\square\\).
(type your answer in interval notation. simplify your answer. use integers or fractions for any numbers in the expression.)
\\(\bigcirc\\) b. the solution set is the empty set.

Explanation:

Step1: Rewrite the inequality

Rewrite \(x^{2}\leq25\) as \(x^{2}-25\leq0\).

Step2: Factor the quadratic

Factor \(x^{2}-25\) using the difference of squares formula \(a^{2}-b^{2}=(a + b)(a - b)\), so we get \((x + 5)(x - 5)\leq0\).

Step3: Find critical points

Set each factor equal to zero: \(x+5 = 0\) gives \(x=-5\) and \(x - 5=0\) gives \(x = 5\). These are the critical points that divide the number line into intervals: \((-\infty,-5)\), \((-5,5)\), and \((5,\infty)\).

Step4: Test intervals

• For the interval \((-\infty,-5)\), let's choose \(x=-6\). Then \((-6 + 5)(-6 - 5)=(-1)\times(-11)=11>0\), so the inequality is not satisfied here.
• For the interval \((-5,5)\), let's choose \(x = 0\). Then \((0 + 5)(0 - 5)=(5)\times(-5)=-25\leq0\), so the inequality is satisfied here.
• For the interval \((5,\infty)\), let's choose \(x = 6\). Then \((6 + 5)(6 - 5)=(11)\times(1)=11>0\), so the inequality is not satisfied here.

Step5: Check critical points

We also need to check the critical points \(x=-5\) and \(x = 5\). When \(x=-5\), \((-5)^{2}=25\leq25\) (true). When \(x = 5\), \(5^{2}=25\leq25\) (true). So the solution includes \(x=-5\) and \(x = 5\).

Answer:

The solution set in interval notation is \([-5,5]\) (so the correct choice is A with the solution set \([-5,5]\)).