The formula this book hands you for the roots, and how to apply it

What to assume they know

• Standard form, and reading a, b and c with their signs — The shape an equation has to have before it counts as quadratic
• Root of an equation and the at-most-two ceiling — Why a root of the equation is the same thing as a zero of the polynomial
• Factorising by splitting the middle term, and its limits — Splitting the middle term, then setting each factor to zero
• Square roots of numbers, including that a square root of a positive number comes in a positive and a negative version
• The identity (m + n)² = m² + 2mn + n², used in reverse
• The angle in a semicircle is a right angle, and the Pythagoras theorem

What they should be able to do

• Write the root formula from memory in the form the chapter prints it, and read each part of it aloud correctly
• Identify a, b and c from an equation in standard form without losing a sign, and substitute them into the formula
• Explain what the ± sign is doing, and produce the two separate roots it stands for
• Derive the formula by multiplying through by 4a and completing a square, and say at which step the requirement on b² − 4ac appears
• State the condition the chapter's Summary attaches to the formula and explain why §4.4's opening line is incomplete without it
• Apply the formula to an equation that arises from a geometric situation, and reject the root the situation cannot use
• Apply the formula to an equation whose discriminant vanishes, and report the single value it returns
• Verify a root obtained from the formula by substitution, and explain why verification matters more when the method has not been proved

Where it usually goes wrong

• "The formula is a shortcut for factorising." It is not a shortcut; it is what you use when no convenient factorisation exists. Exercise 4.3's third item is exactly such a case.
• "± means you choose whichever looks better." It stands for two separate computations, both of which must be carried out. Which of the two answers the question is decided afterwards by the situation, not by preference.
• "c is whatever number you can see." It is the constant term with its sign once the equation is in standard form. A dropped minus sign is the commonest error in this whole chapter.
• "You can use the formula before rearranging." a, b and c only exist once everything is on one side and the powers are in order. Applying the formula to an untidied equation reads the wrong coefficients.
• "The 2a only divides the square-root part." It divides the whole numerator. Writing the formula with a single fraction bar spanning both terms prevents this.
• "Every root the formula returns is an answer." Example 8 returns −12, and a distance cannot be negative. The formula answers the equation; the situation decides which roots survive.
• "If the book gives a formula, it has been proved." In this chapter it has not. That is not a reason to distrust it, but it is a reason to verify results by substitution — which is precisely what the chapter asks after its factorising examples.

Questions to check understanding

• Solve a given quadratic equation by the formula, showing a, b, c and the discriminant as separate steps
• Word problems reaching a quadratic that does not factorise conveniently, where the formula is the only route
• Geometry-based items in which the equation comes from Pythagoras, an area or a perimeter, and one root is then rejected
• Items requiring the answer in surd form rather than as a decimal
• Derivation items asking for the formula to be obtained by completing the square, which appear in the board's question papers even though this edition of the chapter does not print the derivation
• Verification items, where a root obtained from the formula must be substituted back

Examples worth working on the board

Values marked verified are worked out here from the printed data; this chapter prints no answers, and every result below is derived here rather than looked up.

• How §4.4 opens (p. 44). The section begins with a single line introducing the formula and then prints it, with no derivation before or after it. The line as printed is not a grammatical sentence — it reads as though a preceding block of working has been removed. Immediately below, the chapter splits the formula into its two roots for the case where b² − 4ac is positive, writing each as −b⁄2a with the root term added and then subtracted. That split is the whole of what the explanation needs for section 5, and it is printed.
• The condition, and where it is stated. §4.4's opening line carries no proviso. The Summary on p. 47, point 4, restates the formula and does attach one: b² − 4ac must not be negative. Verified by reading both pages as images.
• The derivation to supply (not in the book; not printed anywhere in pp. 38–47). Start from ax² + bx + c = 0 and multiply throughout by 4a, which is not zero, so the roots are unchanged. Verified: that gives 4a²x² + 4abx + 4ac = 0. The first two terms are the opening of the square (2ax + b)², which also carries a b², so the equation becomes (2ax + b)² − b² + 4ac = 0, that is (2ax + b)² = b² − 4ac. Taking square roots gives 2ax + b = ±√(b² − 4ac), and dividing by 2a produces the printed formula. Two things fall out of this and should be pointed at as they appear: the left side is a square, so a real x forces b² − 4ac to be non-negative; and the ± arrives from the square root and nowhere else.
• Reading the formula. Verified: the two roots are −b⁄2a + √(b² − 4ac)⁄2a and −b⁄2a − √(b² − 4ac)⁄2a, exactly as the chapter writes them on p. 44. So the roots sit at equal distances either side of −b⁄2a, and wherever there is a pair to add — that is, whenever the discriminant is not negative — the two sum to −b⁄a. Below zero this chapter reports no roots at all, so there is nothing to total. That last observation is added here, not the chapter's, but it makes a good arithmetic check on any pair of roots.
• Getting a, b and c right. Worth one deliberate beat: in x² + 7x − 60 = 0 the value of c is −60 and not 60, and in 2x² − 4x + 3 = 0 the value of b is −4 and not 4. Verified: using c = 60 in the first would give 49 − 240, a negative discriminant, and the whole verdict would invert. Sign loss here is the single most expensive slip in the chapter.
• Example 8 (§4.4, pp. 45–46, with Fig. 4.2). A pole is to stand somewhere on the rim of a circular park 13 metres across. Two gates, A and B, sit at opposite ends of a diameter. The pole's distances from the two gates must differ by 7 metres. Is such a position possible, and if so where? Verified: taking the distance to gate B as x metres makes the distance to gate A equal to x + 7. Because A and B are the ends of a diameter, the angle the two distances make at the pole is a right angle, so the two distances and the diameter form a right triangle with the diameter as hypotenuse. Pythagoras gives (x + 7)² + x² = 169, which expands to 2x² + 14x − 120 = 0 and halves to x² + 7x − 60 = 0. The discriminant is 49 + 240 = 289, which is positive, so a position exists — and the chapter settles that question before finding the position. The formula then gives x = (−7 ± 17)/2, so 5 or −12. A distance cannot be negative, so the pole stands 5 m from gate B and 12 m from gate A. Verified against the circle: 25 + 144 = 169, and 12 − 5 = 7, so both stated conditions hold.
• Fig. 4.2 (p. 45). A circle with a patterned fill. Gate A sits low on the left of the boundary and gate B high on the right, each marked by a small rectangle drawn across the boundary line — these are gate symbols, not angle marks. The diameter AB runs between them as a diagonal and carries the label 13. The pole P sits on the boundary at the lower right, with the segment from A to P drawn horizontally and the segment from P to B drawn vertically. Read off the printed page and a close-up: the right angle at P is not marked in the figure, and the individual lengths x and x + 7 are not labelled on it either.
• Example 9 (§4.4, p. 46). Equation 3x² − 2x + 1/3 = 0, so a is 3, b is −2 and c is 1/3. Verified: the discriminant is 4 − 4, which is nothing, so the root term contributes nothing and both roots are −b⁄2a, that is 2/6, which is 1/3. Substitution check at 1/3: 1/3 − 2/3 + 1/3 comes to nothing.
• Exercise 4.3 Question 1 (p. 47), three equations to settle and, where real, to solve: 2x² − 3x + 5 = 0; 3x² − 4√3x + 4 = 0; and 2x² − 6x + 3 = 0. Verified: the discriminants are −31, then nothing, then 12; the first has no real root, the second has 2⁄√3 twice, and the third has (3 + √3)/2 and (3 − √3)/2. The third is the exercise's only genuinely irrational pair and is worth working, because factorising will not reach it.

Figures to have open

• Fig. 4.2 (p. 45) redrawn as a schematic, with two additions the printed figure does not carry: the right angle at P marked, and the two distances labelled x and x + 7. Both are needed for the argument and neither is on the page. Do not reproduce the printed artwork.
• A formula annotation card, with the numerator's two terms, the bar and the denominator each callable out separately. Standard schematic.
• A derivation board for section 4, holding all five lines at once so the appearance of b² − 4ac as the value of a square is visible. Standard schematic, and not in the book — this working is nowhere in the chapter.
• A centre-and-spread number line: −b⁄2a marked, with equal arrows either side. Standard schematic, an added device.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, Chapter 4 "Quadratic Equations", §4.4 Nature of Roots, p. 44 — the opening line, the formula, and its split into two roots for the positive case
• Same chapter, §4.4, pp. 45–46 — Example 8 with Fig. 4.2, and Example 9
• Same chapter, §4.5 Summary, p. 47, point 4 — the formula restated with the condition on b² − 4ac attached
• Same chapter, §4.1, p. 39 — the historical note naming completing the square and crediting Sridharacharya, quoted by Bhaskara II
• Same chapter, Exercise 4.3 Question 1, p. 47
• Companion topic Why the sign of b² − 4ac settles how many real roots exist owns the three-case reading of the discriminant; this topic uses the discriminant only as a number to compute