Estimating square roots

Every whole number $x$ that is not a perfect square sits somewhere between two consecutive perfect squares, $a^2$ and $b^2 = (a+1)^2$. Because squaring is an increasing operation for positive numbers, taking square roots preserves the order:

where: $a$ and $b=a+1$ are consecutive whole numbers, $x$ is the number whose root you want, and $a^2$, $b^2$ are the perfect squares bracketing it. This single line already answers the first, coarsest kind of question — "between which two whole numbers does $\sqrt{70}$ lie" — without any arithmetic beyond recalling that $8^2=64$ and $9^2=81$. The bracket by itself is often too rough for classwork, which usually asks for one decimal place. The fix is to notice that $\sqrt{x}$ does not sit exactly halfway between $a$ and $b$ just because $x$ sits between $a^2$ and $b^2$ — it sits closer to whichever end $x$ is closer to. Comparing the two gaps, $x-a^2$ and $b^2-x$, tells you which way to lean. Pushing that idea into a formula gives a simple linear estimate:

where: $x-a^2$ is how far $x$ has climbed above the lower perfect square, and $b^2-a^2$ is the full gap between the two perfect squares, so the fraction is "how far across the gap" $x$ sits, added on to $a$. This is not exact — the true relationship between $x$ and $\sqrt{x}$ is a curve, not a straight line — but for a one-decimal classroom estimate it is close enough, and it is always checked afterwards by squaring the estimate back. Notice, too, why estimation matters as a habit and not just a technique. Suppose a long-division calculation for $\sqrt{70}$ came out to $87$ instead of $8.37$ because of a misplaced decimal. The bracket $8<\sqrt{70}<9$, found in seconds from perfect squares alone, would flag that answer as absurd before you even checked the working line by line. Estimation is therefore a fast first pass that catches large errors early, and it is also the intuition you carry forward when, in later classes, you place irrational numbers such as $\sqrt2$ or $\sqrt3$ on a number line. The figure below shows why the linear guess works reasonably well over a short stretch: between two consecutive perfect squares, the curve $y=\sqrt{x}$ is nearly straight, so sliding along the chord from $(a^2,a)$ to $(b^2,b)$ lands close to the true curve.

Estimate $\sqrt{70}$ to one decimal place, then use the same method on $\sqrt{200}$.

Step 1. Bracket $70$: we know $8^2=64$ and $9^2=81$, and $64<70<81$, so $8<\sqrt{70}<9$.

Step 2. Compare the gaps: $70-64=6$ and $81-70=11$. Since $6<11$, $70$ is closer to $64$, so $\sqrt{70}$ leans toward $8$, not $9$ — a first guess around $8.3$–$8.4$.

Step 3. Linear estimate: $\sqrt{70}\approx 8+\dfrac{70-64}{81-64}=8+\dfrac{6}{17}\approx 8+0.35=8.35$.

Step 4. Verify by trial squaring: $8.3^2=68.89$ (too low), $8.4^2=70.56$ (too high). Since $70$ sits between these two squares, the estimate of about $8.4$ is confirmed as reasonable, and the check also shows the true value must lie between $8.3$ and $8.4$. $\sqrt{70}\approx 8.4$ (the true value to two decimal places is $8.37$). This four-step routine is worth practising until it is automatic: bracket, compare gaps, apply the fraction, then square-check. The check in Step 4 is not optional busywork — it is what turns a guess into a defensible estimate, because it independently confirms the bracket without relying on the linear formula being exact.

Competitive stretch

The same four steps on $x=200$: $14^2=196$ and $15^2=225$ bracket it, so $14<\sqrt{200}<15$. The gaps are $200-196=4$ and $225-200=25$, so $200$ sits very close to $196$, and the linear estimate gives $14+\dfrac{4}{29}\approx 14.14$. Checking, $14.1^2=198.81$ and $14.2^2=201.64$ bracket $200$, confirming it. $\sqrt{200}\approx 14.1$, between the whole numbers $14$ and $15$.

• Rounding to the nearer perfect square instead of actually comparing $x-a^2$ with $b^2-x$ — a number can look "close to" the bigger square by eye and still be numerically closer to the smaller one.
• Treating the linear-estimate answer as exact — it is an approximation good to about one decimal place; only long division (within the syllabus's digit cap) gives the precise value.
• Skipping the trial-squaring check — this one step catches most arithmetic slips immediately and costs almost no extra time.
• Confusing "estimate this square root" with "prove this number is irrational" — Class 8 asks only for a numerical estimate; classifying $\sqrt2$, $\sqrt3$, etc. as irrational is Class 9 content and should not be mixed in here.

Ask the tutor: Not sure which two perfect squares to bracket your number between — ask the AI tutor to walk through estimating any square root step by step.