CodeVerse Creatives
Esc
Numerical Aptitude Syllabus / Module 10 · Equations / Week 10

Week 10: Linear & Quadratic Equations

Every word problem this course throws at you eventually becomes an equation — this week builds the skill of translating text into algebra reliably, plus the fast factoring method and root-relationship shortcuts that keep quadratics from turning into a formula-plugging exercise.

Module 10 of 20 Week 10 of 20 ~2–3 Hours Practice Problems Included

By the end of this week, you'll be able to

  • Translate a word problem into a linear equation systematically
  • Factor a quadratic quickly using the sum-and-product method
  • Find the sum and product of a quadratic's roots without solving it
1 Translating Word Problems 2 Factoring Quadratics Fast 3 Sum & Product of Roots 4 The Discriminant & Nature of Roots 5 Practice Problems 6 Knowledge Check

1. Translating Word Problems

The actual skill in "algebra word problems" is translation, not solving — once the equation is set up correctly, the algebra is usually simple. A consistent process removes most of the guesswork.

the translation process
1. Identify what's UNKNOWN -- name it x (or the smallest
   number of unknowns possible)
2. Express every OTHER quantity in terms of x
3. Find the ONE sentence in the problem that states an
   EQUALITY -- that's your equation
4. Solve, then check the answer against the ORIGINAL
   word problem (not just the equation)

Worked: "A father is 3 times as old as his son. In 12 years,
he will be twice as old. Find their current ages."

1. Unknown: son's current age = x
2. Father's current age = 3x
3. Equality: in 12 years, father = 2 x son
   (3x + 12) = 2(x + 12)
4. Solve: 3x+12 = 2x+24 -> x = 12
   Son = 12, Father = 36
   Check: in 12 years, son=24, father=48 -- 48 = 2x24 -- correct

Step 4's check against the original wording (not just the equation) catches translation errors specifically — a wrong equation can still be solved "correctly" and produce a wrong final answer, so verifying against the actual problem statement is the only way to catch that class of mistake.

2. Factoring Quadratics Fast

For a quadratic x² + bx + c = 0, factoring is faster than the quadratic formula whenever two integers can be found that multiply to c and add to b — worth trying first, every time, before reaching for the formula.

sum-and-product factoring
Solve x^2 - 7x + 12 = 0

Need two numbers that MULTIPLY to 12 and ADD to -7:
  -3 and -4:  (-3)x(-4)=12 correct, (-3)+(-4)=-7 correct

x^2 - 7x + 12 = (x-3)(x-4) = 0
x = 3 or x = 4

For a leading coefficient other than 1 (e.g. 2x² + 7x + 3), the same idea extends: find two numbers multiplying to a×c (here 2×3=6) and adding to b (7) — here, 1 and 6 — then split the middle term using those numbers and factor by grouping.

3. Sum & Product of Roots

A very common test-question shape asks for something about a quadratic's roots (like their sum, or the sum of their squares) without asking you to find the roots themselves — which is exactly what this shortcut is built for.

for ax^2 + bx + c = 0
Sum of roots     = -b/a
Product of roots =  c/a

Worked: for 2x^2 - 5x + 3 = 0, find the sum and product of
roots WITHOUT solving.

  Sum = -(-5)/2 = 5/2
  Product = 3/2

Verify by actually factoring: 2x^2-5x+3 = (2x-3)(x-1) = 0
  Roots: x=3/2, x=1
  Sum: 3/2+1 = 5/2 -- matches
  Product: 3/2 x 1 = 3/2 -- matches

This unlocks fast answers to questions like "find the sum of the squares of the roots": since (sum)² = (sum of squares) + 2×(product), you can solve for the sum of squares directly from the sum and product alone, never touching the individual root values at all.

4. The Discriminant & Nature of Roots

Before solving a quadratic, the discriminant tells you what kind of roots to expect — useful for eliminating answer options quickly, or catching a setup error if the expected answer type doesn't match.

discriminant D = b^2 - 4ac
D > 0:  two distinct real roots
D = 0:  exactly one real root (a repeated root)
D < 0:  no real roots (roots are complex, not testable
        on most aptitude exams -- signals a likely error
        if you expected a real-number answer)

Worked: for x^2 - 4x + 4 = 0
  D = (-4)^2 - 4(1)(4) = 16 - 16 = 0  -> one repeated root
  Factoring confirms: (x-2)^2 = 0 -> x = 2 (repeated)
A negative discriminant is a red flag on a word problem

If a quadratic derived from a real-world word problem (ages, speeds, quantities) has a negative discriminant, that's a strong signal the equation was set up incorrectly — real physical quantities should produce real roots. Checking the discriminant before solving can catch a translation mistake before you invest time solving an equation that was never going to give a sensible answer.

5. Practice Problems

Practice

30 problems, no calculator — Easy, Medium & Tough

Try sum-and-product factoring before reaching for the quadratic formula. Easy problems should take under 30 seconds each; Medium under 60 seconds; Tough under 90 seconds.

Easy (1–10)

  1. Solve: x + 5 = 12.
  2. Solve: 3x = 21.
  3. Solve: 2x - 4 = 10.
  4. Solve: x/3 = 9.
  5. Solve: 5x + 2 = 22.
  6. Factor and solve: x² - 9 = 0.
  7. Factor and solve: x² - 5x = 0.
  8. Solve: x² - 7x + 12 = 0.
  9. Find the sum of the roots of x² - 6x + 8 = 0 without solving.
  10. Find the product of the roots of x² - 6x + 8 = 0 without solving.
Easy — Answers

1) x=7. 2) x=7. 3) x=7. 4) x=27. 5) x=4. 6) x=±3. 7) x=0 or 5. 8) (x-3)(x-4)=0 → x=3,4. 9) Sum=6. 10) Product=8.

Medium (11–20)

  1. Solve: 2x + 3 = x + 9.
  2. The sum of two numbers is 25 and their difference is 7. Find the numbers.
  3. Solve: x² - 3x - 10 = 0.
  4. Solve: 2x² - 7x + 3 = 0.
  5. A number is 5 more than twice another. Their sum is 29. Find the numbers.
  6. Find the discriminant of x² + 4x + 4 = 0 and state the nature of its roots.
  7. Solve: x² + x - 12 = 0.
  8. If the sum of a number and its reciprocal is 26/5, find the number(s).
  9. The sum of the digits of a two-digit number is 9. If the digits are reversed, the number increases by 27. Find the number.
  10. Solve for x: 3(x-2) = 2(x+1).
Medium — Answers

11) x=6. 12) x=16, y=9. 13) (x-5)(x+2)=0 → x=5,-2. 14) (2x-1)(x-3)=0 → x=1/2, 3. 15) 2y+5+y=29 → y=8, first number=21. 16) D=16-16=0 → equal roots (repeated x=-2). 17) (x+4)(x-3)=0 → x=-4,3. 18) 5x²-26x+5=0 → (5x-1)(x-5)=0 → x=5 or 1/5. 19) a+b=9, b-a=3 → a=3,b=6 → number=36. 20) 3x-6=2x+2 → x=8.

Tough (21–30)

  1. A father's age is 4 times his son's age. In 16 years, he will be twice as old. Find their current ages.
  2. Find the sum of the squares of the roots of x² - 5x + 6 = 0 without solving for the roots directly.
  3. Solve: x² - 2x - 15 = 0, and verify using the sum and product of roots.
  4. The sum of a number and its square is 132. Find the number.
  5. Two numbers differ by 3, and the sum of their squares is 89. Find the numbers.
  6. A rectangular field's length is 3m more than its width. If the area is 154 sq.m, find the dimensions.
  7. Find the value of k such that x² - kx + 36 = 0 has equal roots.
  8. The denominator of a fraction is 3 more than the numerator. If 2 is added to both, the fraction becomes 4/5. Find the original fraction.
  9. A sum of Rs. 720 was divided among A, B and C such that A got Rs. 20 more than B, and B got Rs. 20 more than C. Find each person's share.
  10. Solve: x² - (a+b)x + ab = 0, in terms of a and b.
Tough — Answers

21) F=4S; F+16=2(S+16) → 2S=16 → S=8, F=32. 22) Sum=5, Product=6; sum of squares=5²-2×6=13. 23) (x-5)(x+3)=0 → x=5,-3; sum=2, product=-15, both match -b/a and c/a. 24) x²+x-132=0 → (x+12)(x-11)=0 → x=11. 25) x-y=3, xy=40 (from (x-y)²=x²+y²-2xy), x+y=13 → x=8, y=5. 26) w(w+3)=154 → w=11, length=14. 27) Equal roots: k²=4×36 → k=12. 28) (x+2)/(x+5)=4/5 → x=10 → fraction=10/13. 29) C=x, B=x+20, A=x+40; 3x+60=720 → x=220 → C=220, B=240, A=260. 30) (x-a)(x-b)=0 → x=a or b.

6. Knowledge Check

Four quick questions. Expand each to check your answer.

Q1

Why is checking a word-problem answer against the original text more important than just re-checking the algebra?

Re-checking the algebra only confirms the equation was solved correctly — it can't catch a translation error, where the equation itself doesn't correctly represent what the word problem actually described. Checking the numeric answer against the original sentences is the only step that catches a wrong equation that was nonetheless solved perfectly.

Q2

For x² - 7x + 12 = 0, why do the two factoring numbers need to multiply to 12 and add to -7, specifically?

Expanding (x-p)(x-q) gives x² - (p+q)x + pq — matching this against x² - 7x + 12 term by term requires p+q = 7 (giving -7x when negated) and pq = 12. The two target numbers are exactly the sum-and-product structure that makes the expanded factored form match the original quadratic's coefficients.

Q3

Why can the sum of squares of a quadratic's roots be found without ever solving for the individual roots?

The algebraic identity (sum)² = (sum of squares) + 2×(product) relates the sum of squares directly to the sum and product of roots, both of which come straight from the -b/a and c/a shortcuts. Since all three quantities in that identity are known except the sum of squares, it can be isolated algebraically without ever computing the individual root values.

Q4

Why is a negative discriminant on a quadratic derived from a real-world word problem a signal to double-check the setup?

A negative discriminant means the equation has no real number solutions, but a word problem about ages, speeds, or quantities is describing something that genuinely exists in the real world and must have a real numeric answer. If the derived equation produces no real roots, that's strong evidence the equation itself was set up incorrectly during translation, not that the problem has no answer.

← Back to Week 9: Time, Speed & Distance Up next Week 11: Progressions — AP, GP & Series →