QuanticaPrealgebraOpen in the courseSquaring a side gives the area, so \(5^2 = 25\). This lesson flips the question. Given that a panel uses exactly 36 tiles in a square, how long was one edge? That backward move, recovering the side from the area, is the square root.
Problem
A square concrete pad covers \(196\) square feet. Find the length of one edge in feet.
Show a hint
- Squaring a side gives the area, so to get the side back you undo the square. You are hunting for the number that, times itself, lands exactly on \(196\).
- The two landmarks \(13 \times 13 = 169\) and \(15 \times 15 = 225\) trap the answer between \(13\) and \(15\). Only one whole number lives in that gap, so test it by multiplying it by itself.
Show the full solution
\(13^2 = 169\) falls short and \(15^2 = 225\) overshoots, so the side is the one whole number in between. Check it, \(14 \times 14 = 196\), so one edge is \(\boxed{14}\) feet. Squaring a side gives the area, so rooting the area gives the side back.
Problem
A video wall is built from \(625\) identical square panels in a perfect square. Since \(20^2 = 400\) and \(30^2 = 900\), the side length is in the twenties. What is \(\sqrt{625}\)?
Show a hint
- The side length is a number in the twenties whose square is \(625\). Try squaring a value right in the middle of that range and see how close you land.
- Test \(25\). Compute \(25 \times 25\). If it equals \(625\) exactly, then \(25\) is the side length, since the square root recovers the side of a square from its area.
Show the full solution
The side is in the twenties, so try the middle of that range. \(25 \times 25 = 625\) exactly, so \(\sqrt{625} = \boxed{25}\) panels along one side. Squaring two landmarks first pins a root into a small range, which leaves only a guess or two to test.
The radical \(\sqrt{36}\) names a single value, \(6\), the nonnegative side of a square with area \(36\). The equation \(x^2 = 36\) asks something different. It wants every number whose square is \(36\), and there are two of those, \(6\) and \(-6\). The two questions look alike on the page, so read carefully to tell which one a problem is handing you.
Problem
Both \(10\) and \(-10\) square to \(100\), so two numbers satisfy \(x^2 = 100\). But the radical \(\sqrt{100}\) names only one of them, the nonnegative one. Evaluate \(\sqrt{100}\).
Show a hint
- The radical does not mean "any number that squares to 100". It means the one nonnegative number whose square is 100. Of the two candidates 10 and \(-10\), only one is zero or positive.
- Ask which value is nonnegative. Since \(10^2 = 100\) and 10 is positive, the radical names 10, not \(-10\).
Show the full solution
\(10^2 = 100\) and \(10\) is nonnegative, so \(\sqrt{100} = \boxed{10}\). Both \(10\) and \(-10\) square to \(100\), but the radical is defined to name only the nonnegative one, so it never reports \(-10\).
Problem
The equation \(x^2 = 49\) asks for every number whose square is \(49\). Since \(7^2 = 49\) and \((-7)^2 = 49\), both \(7\) and \(-7\) work. How many values of \(x\) satisfy \(x^2 = 49\)?
Show a hint
- Squaring a negative number gives a positive result, so do not stop at the first number you find. Try both a positive and a negative candidate and square each one.
- Check \(7^2\) and check \((-7)^2\). If both land on \(49\), then count how many separate values of \(x\) you have found.
Show the full solution
\(7^2 = 49\), and since a negative times a negative is positive, \((-7)^2 = 49\) as well. Nothing else squares to \(49\), so exactly \(\boxed{2}\) values work, \(7\) and \(-7\). The equation keeps both, while \(\sqrt{49}\) reports only the nonnegative \(7\).
Problem
A depth sensor satisfies \(x^2 = 324\). Two values work: \(x = 18\) and \(x = -18\). Since depth is negative, which value is \(x\)?
Show a hint
- Find the two numbers whose square is \(324\). Take the square root of \(324\) to get the size, then remember the negative version squares to \(324\) as well.
- The size of \(x\) is \(\sqrt{324} = 18\), so \(x\) is \(18\) or \(-18\). Since \(x\) measures a depth below the surface, it must be negative, which rules out \(18\).
Show the full solution
Since \(18^2 = 324\), the two numbers that square to \(324\) are \(18\) and \(-18\). A depth below the surface is negative, so \(x = \boxed{-18}\). Squaring erases the sign, so the equation alone leaves both candidates and the story decides which one you keep.
Problem
Squaring any real number gives a result that is zero or positive. So how many real values of \(x\) satisfy \(x^2 = -9\)?
Show a hint
- Try a few numbers. \(3^2 = 9\), \((-3)^2 = 9\), \(0^2 = 0\). Every square you get is \(0\) or positive. Can any of them ever equal \(-9\)?
- Since squaring any real number gives a result of \(0\) or more, no real number can square to a negative like \(-9\). Count how many real \(x\) work, and that count is the answer.
Show the full solution
A positive squared is positive, a negative squared is positive, and \(0^2 = 0\). That covers every real number, so a square is always \(0\) or greater and nothing can reach \(-9\). The count of real solutions is \(\boxed{0}\).
Problem
Evaluate \(\sqrt{(-12)^2}\). Square first, then take the root. What whole number do you get?
Show a hint
- Follow the order of operations and square first. What is \((-12)^2\)? Remember that a negative times a negative gives a positive result, so the inside becomes a positive number before you ever touch the radical.
- Once the inside is \(144\), you just need \(\sqrt{144}\). Find the nonnegative number whose square is \(144\). The radical always reports the nonnegative root, so the sign you started with does not survive.
Show the full solution
Square first. \((-12)^2 = 144\), and \(\sqrt{144} = 12\), so the value is \(\boxed{12}\). Squaring drops the sign and the radical returns the nonnegative root, which is why the answer comes out positive even though the number inside started negative.
Problem
A square garden covers \(1{,}296\) sq ft. Notice \(1{,}296 = 16 \times 81\). Use \(\sqrt{1{,}296} = \sqrt{16} \times \sqrt{81}\) to find the side length.
Show a hint
- Split the work using \(1{,}296 = 16 \times 81\). You can take the root of each piece on its own, since \(\sqrt{16 \times 81} = \sqrt{16} \times \sqrt{81}\).
- Find each small root, then multiply them. \(\sqrt{16} = 4\) because \(4^2 = 16\), and \(\sqrt{81} = 9\) because \(9^2 = 81\). Now compute \(4 \times 9\).
Show the full solution
Since \(1{,}296 = 16 \times 81\), the root splits, so \(\sqrt{1{,}296} = \sqrt{16} \times \sqrt{81} = 4 \times 9 = 36\). Checking, \(36 \times 36 = 1{,}296\), so the side is \(\boxed{36}\) feet. Splitting into perfect-square factors turns one hard root into two easy ones.
Problem
A classmate claims \(\sqrt{1024} = 34\). But \(34^2 = 1156 > 1024\), and a number ending in 4 squares to something ending in 6, not 4. What is the correct value of \(\sqrt{1024}\)?
Show a hint
- The size test says the root sits below 34, and the last-digit test says it cannot end in 4. So try a slightly smaller whole number and square it to see if you hit \(1{,}024\) on the nose.
- You want a number whose square ends in 4. A number ending in 2 works, since \(2 \times 2 = 4\). Check \(32^2\).
Show the full solution
The root sits below \(34\), and it has to end in \(2\), since only a number ending in \(2\) squares to something ending in \(4\). That points at \(32\), and \(32 \times 32 = 1{,}024\) exactly, so \(\sqrt{1024} = \boxed{32}\). The size check and the last-digit check together usually leave a single candidate to test.
Problem
A square herb bed has a side of \(15\) feet, so \(\sqrt{x} = 15\) where \(x\) is the area. Square both sides to free \(x\). What is \(x\)?
Show a hint
- The variable is stuck under the radical sign. Whatever you do to one side of an equation you must do to the other, so apply the same operation to both sides that will cancel the square root.
- Squaring undoes a square root. The left side \((\sqrt{x})^2\) collapses back to just \(x\), and the right side becomes \(15^2\). Now just compute \(15^2\).
Show the full solution
Square both sides. $$\left(\sqrt{x}\right)^2 = 15^2$$ The left side collapses to \(x\), and \(15^2 = 225\), so \(x = \boxed{225}\) square feet. Squaring is the move that frees a variable from under a radical, and the check holds, since \(\sqrt{225} = 15\).
Problem
Solve \(\sqrt{3x + 4} = 11\). Square both sides, then solve the resulting linear equation. What is \(x\)?
Show a hint
- Squaring is the move that undoes a square root. If \(\sqrt{3x + 4} = 11\), then squaring both sides gives \(3x + 4 = 11^2\). Work out \(11^2\) first.
- Now \(3x + 4 = 121\) is just a linear equation like the ones from Chapter 6. Subtract \(4\) from both sides, then divide by \(3\).
Show the full solution
Squaring both sides gives \(3x + 4 = 11^2 = 121\). Subtract \(4\) to get \(3x = 117\), then divide by \(3\), so \(x = \boxed{39}\). Checking, \(3 \times 39 + 4 = 121\) and \(\sqrt{121} = 11\). Squaring clears the radical and leaves an ordinary linear equation.
Problem
How many solutions does \(\sqrt{x - 2} = -4\) have?
Show a hint
- The radical \(\sqrt{\phantom{x}}\) always means the nonnegative root. So whatever \(x - 2\) turns out to be, can \(\sqrt{x - 2}\) ever come out negative?
- Try the tempting answer anyway. Squaring gives \(x = 18\), and then \(\sqrt{18 - 2} = \sqrt{16} = 4\), not \(-4\). So even that candidate fails the check. If the one value squaring hands you does not work, count how many values are left.
Show the full solution
The radical always returns a value that is zero or positive, and the right side is \(-4\), so nothing can match. Squaring hands you \(x - 2 = 16\) and \(x = 18\), but \(\sqrt{18 - 2} = \sqrt{16} = 4\), not \(-4\), so that candidate fails too. The number of solutions is \(\boxed{0}\). Squaring can produce answers that do not satisfy the original equation, which is why you check every candidate back.
Practice these ideas
Practice
A square garden bed covers \(121\) sq ft. What is the side length, in feet? Evaluate \(\sqrt{121}\).
Show the solution
\(10^2 = 100\) is too small and \(12^2 = 144\) is too big, so try \(11\). Since \(11 \times 11 = 121\), the bed is \(\boxed{11}\) feet on a side.
Practice
A square rug covers \(256\) sq ft. What is its side length in feet?
Show the solution
The side length is \(\sqrt{256}\). Since \(15^2 = 225\) is too small and \(17^2 = 289\) is too big, test \(16\), and \(16 \times 16 = 256\). The rug is \(\boxed{16}\) feet on a side.
Practice
A square patio covers \(169\) sq ft. Since \(10^2 = 100\) and \(20^2 = 400\), the side length is between \(10\) and \(20\). Find \(\sqrt{169}\).
Show the solution
The root is between \(10\) and \(20\), and since \(169\) ends in \(9\) the root must end in \(3\) or \(7\). Testing the smaller option, \(13 \times 13 = 169\), so each side of the patio is \(\boxed{13}\) feet.
Practice
When you write \(x^2 = 9\), both \(3\) and \(-3\) work. But \(\sqrt{9}\) names only the nonnegative one. Evaluate \(\sqrt{9}\).
Show the solution
\(3 \times 3 = 9\) and \(3\) is nonnegative, so \(\sqrt{9} = \boxed{3}\). The equation \(x^2 = 9\) keeps both \(3\) and \(-3\), but the radical names only the nonnegative one.
Practice
A square garden covers \(400\) sq ft. What is \(\sqrt{400}\)?
Show the solution
Split into perfect-square factors, since \(400 = 4 \times 100\). Then \(\sqrt{400} = \sqrt{4} \times \sqrt{100} = 2 \times 10 = 20\), and squaring back gives \(20^2 = 400\). The garden is \(\boxed{20}\) feet on a side.
Practice
Both \(9\) and \(-9\) square to \(81\), since \(9^2 = 81\) and \((-9)^2 = 81\). How many values of \(x\) satisfy \(x^2 = 81\)?
Show the solution
\(9^2 = 81\), and a negative times a negative is positive, so \((-9)^2 = 81\) as well. Nothing else squares to \(81\), so exactly \(\boxed{2}\) values work, \(9\) and \(-9\). Solving \(x^2 = 81\) gives a pair, while evaluating \(\sqrt{81}\) gives just \(9\).
Practice
The equation \(x^2 = 225\) has two solutions, \(15\) and \(-15\). Engineers know \(x\) is negative. What is \(x\)?
Show the solution
Since \(15^2 = 225\), both \(15\) and \(-15\) satisfy \(x^2 = 225\). The engineers know \(x\) is negative, so \(x = \boxed{-15}\). The equation keeps both roots, and the context tells you which one to report.
Practice
Squaring any real number gives zero or a positive result, never negative. How many real values of \(x\) satisfy \(x^2 = -16\)?
Show the solution
A positive squared is positive, a negative squared is positive, and \(0^2 = 0\), so \(x^2\) is never negative for a real \(x\). Nothing can square to \(-16\), so the count of real solutions is \(\boxed{0}\).
Practice
Squaring undoes a square root and vice versa. Evaluate \(\left(\sqrt{47}\right)^2\) without finding the decimal value of \(\sqrt{47}\).
Show the solution
\(\sqrt{47}\) is the nonnegative number whose square is \(47\), so squaring it hands \(47\) straight back, giving \(\left(\sqrt{47}\right)^2 = \boxed{47}\). No decimal is needed, since squaring and the square root undo each other for any nonnegative number.
Practice
A recorded side length of \(-8\) has the wrong sign. Evaluate \(\sqrt{(-8)^2}\) to recover the nonnegative side length.
Show the solution
Square first, \((-8)^2 = 64\), then take the root, \(\sqrt{64} = \boxed{8}\). Squaring drops the sign and the radical returns the nonnegative value, so \(\sqrt{x^2}\) gives the size rather than the original \(-8\).
Practice
A square hall uses \(4{,}900\) tiles. Since \(4{,}900 = 49 \times 100\), use \(\sqrt{4{,}900} = \sqrt{49} \times \sqrt{100}\) to find the number of tiles along one edge.
Show the solution
Since \(4{,}900 = 49 \times 100\), the root splits, so \(\sqrt{4{,}900} = \sqrt{49} \times \sqrt{100} = 7 \times 10 = 70\). Checking, \(70^2 = 4{,}900\), so one edge of the hall holds \(\boxed{70}\) tiles.
Practice
A friend claims \(\sqrt{289} = 17\). Does \(17^2 = 289\)? What is \(\sqrt{289}\)?
Show the solution
\(17 \times 17 = 289\) exactly, so the friend is right and \(\sqrt{289} = \boxed{17}\). The last-digit clue agrees, since a square ending in \(9\) can only come from a root ending in \(3\) or \(7\).
Practice
You know \(\sqrt{x} = 19\). Square both sides to find \(x\). What is \(x\)?
Show the solution
Square both sides. The left side gives back \(x\), and the right gives \(19^2 = 361\), so \(x = \boxed{361}\). Checking the original, \(\sqrt{361} = 19\) since \(19 \times 19 = 361\).
Practice
A game requires solving \(\sqrt{x + 7} = -3\). The radical is always nonnegative. How many solutions exist?
Show the solution
The radical is always zero or positive, and \(-3\) is negative, so no value of \(x\) can make the two sides match. The count is \(\boxed{0}\). Squaring would give \(x + 7 = 9\) and \(x = 2\), but \(\sqrt{2 + 7} = 3\), not \(-3\), so even that candidate fails the check.