0063-MWALA-LEARN-QUADRATIC-EQUATIONS-SOLVING-FOREVER-2

Objectives: QUADRATIC-EQUATIONS-SOLVING-FOREVER-2

Quadratic Equations Questions

Quadratic Equations Practice

1. Solve the following quadratic equations by the factorization method:

  1. xΒ² + 3x = 0
  2. 3xΒ² βˆ’ 15x = 0
  3. x = 3xΒ²
  4. 2xΒ² = 3x
  5. x(5 βˆ’ x) = 0
  6. 7xΒ² βˆ’ 3x = 0
  7. xΒ² + 3x βˆ’ 40 = 0
  8. 3xΒ² βˆ’ 7x βˆ’ 6 = 0
  9. 12xΒ² + 13x + 3 = 0
  10. xΒ² + 3x + 2 = 0
  11. xΒ² βˆ’ 10x + 24 = 0
  12. 2xΒ² βˆ’ x βˆ’ 6 = 0
  1. 3x + 2 = 9xΒ²
  2. βˆ’3xΒ² + 11x βˆ’ 10 = 0
  3. 4xΒ² = 25
  4. yΒ² βˆ’ 36 = 0
  5. (x βˆ’ 8)Β² = 36
  6. 4xΒ² = 20x βˆ’ 25
  7. 9yΒ² βˆ’ 6y + 1 = 0
  8. (x + 3)Β² βˆ’ 49 = 0

2. Solve the following quadratic equations by completing the square:

  1. xΒ² + 6x + 7 = 0
  2. xΒ² βˆ’ 11x + 1 = 0
  3. xΒ² βˆ’ 7x βˆ’ 7 = 0
  4. 2xΒ² βˆ’ 10x + 7 = 0
  5. mΒ² + 5m = 1
  1. pΒ² βˆ’ 10p + 5 = 0
  2. 2bΒ² = 8b + 11
  3. 3aΒ² βˆ’ 12a = 2
  4. 5nΒ² = 20n + 28
  5. cΒ² βˆ’ 8c + 13 = 0
Math Questions

Mathematics Questions

3. Use the quadratic formula to solve the following equations:
  • (a) 4xΒ² βˆ’ 4x + 1 = 0
  • (b) 5xΒ² + 12x + 3 = 0
  • (c) (3x βˆ’ 2)(2x βˆ’ 5) = 5x(x βˆ’ 2)
  • (d) 3xΒ² = 4x + 4
  • (e) 2xΒ² βˆ’ 5x + 2 = 0
  • (f) 5xΒ² βˆ’ x βˆ’ 18 = 0
4. A man is 4 times as old as his son. In 4 years the product of their ages will be 520. Find the son's present age.
5. Sada has 1,800 shillings to buy pencils. There are two types of pencils whose prices differ by 40 shillings. If she buys the cheaper type, she will get 12 more pencils than if she buys the expensive type. What are the prices of the two types of pencils?
6. Find two consecutive numbers such that the sum of their squares is equal to 145.
7. A picture measures 12 cm by 9 cm and is surrounded by a frame whose area is 180 square centimeters. Find the value of x as shown in the figure.
x
x
x
x
12 cm
9 cm
8. When 6 is divided by a certain number, the result is the same as when 5 is added to the number and that sum divided by 6. Find the number.
9. Find the whole number such that twice its square is 11 more than 21 times the number.
10. A piece of wire 56 cm long is bent to form a rectangle of area 171 cmΒ². Find the dimensions of the rectangle.
Quadratic Equations – Full, Beginner-Friendly Solutions

Quadratic Equations – Step-by-Step Answers

Quick Toolbox (Symbols & Ideas)
  • = β€œequals” – left side has the same value as the right side.
  • β†’ β€œleads to / gives” – the next step logically follows.
  • Β± β€œplus or minus” – means there are two values: one with + and one with βˆ’. We use this when we take square roots: if u2 = k, then u = ±√k.
  • √ β€œsquare root” – a number which squared gives the inside number (e.g., √9 = 3).
  • Zero-Product Property: if AB = 0, then either A = 0 or B = 0. (Basic property of real numbers; used in factorization.)
  • Completing the Square: turn x2 + bx + c into a perfect square like (x + b/2)2 plus/minus a number. History: The method was systematized by Al-Khwarizmi (~820 CE) in his book Al-Jabr (origin of the word β€œalgebra”). Earlier versions were used by Babylonian mathematicians.
  • Difference of squares: a2 βˆ’ b2 = (a βˆ’ b)(a + b). This identity appears in ancient Greek mathematics (e.g., Euclid).
Real-life where quadratics appear
  • Projectiles: height vs. time of a thrown ball is quadratic.
  • Areas: changing length/width of rectangles leads to quadratic expressions.
  • Business: profit = revenue βˆ’ cost; with price/quantity relations, profit often becomes quadratic.

1) Solve by factorization (use Zero-Product Property)

(a) x2 + 3x = 0

  1. Factor out x: x(x + 3) = 0
  2. By Zero-Product Property β†’ x = 0 or x + 3 = 0
  3. So, x = 0 or x = βˆ’3

(b) 3x2 βˆ’ 15x = 0

  1. Factor 3x: 3x(x βˆ’ 5) = 0
  2. β†’ 3x = 0 or x βˆ’ 5 = 0
  3. x = 0 or x = 5

(c) x = 3x2

  1. Rearrange to 0: 3x2 βˆ’ x = 0
  2. Factor x: x(3x βˆ’ 1) = 0
  3. β†’ x = 0 or 3x βˆ’ 1 = 0 β‡’ x = 1/3

(d) 2x2 = 3x

  1. Bring all to 0: 2x2 βˆ’ 3x = 0
  2. Factor x: x(2x βˆ’ 3) = 0
  3. β†’ x = 0 or x = 3/2

(e) x(5 βˆ’ x) = 0

  1. Already factored: x = 0 or 5 βˆ’ x = 0 β‡’ x = 5

(f) 7x2 βˆ’ 3x = 0

  1. Factor x: x(7x βˆ’ 3) = 0
  2. β†’ x = 0 or x = 3/7

(g) x2 + 3x βˆ’ 40 = 0

  1. Find numbers with product βˆ’40 and sum 3 β†’ 8 and βˆ’5
  2. Factor: (x + 8)(x βˆ’ 5) = 0
  3. β†’ x = βˆ’8 or x = 5

(h) 3x2 βˆ’ 7x βˆ’ 6 = 0

  1. Split middle: βˆ’7x β†’ βˆ’9x + 2x (since 3Β·(βˆ’6)=βˆ’18; βˆ’9 + 2 = βˆ’7)
  2. Group: (3x2 βˆ’ 9x) + (2x βˆ’ 6) = 3x(x βˆ’ 3) + 2(x βˆ’ 3)
  3. Factor: (3x + 2)(x βˆ’ 3) = 0
  4. β†’ x = 3 or x = βˆ’2/3

(i) 12x2 + 13x + 3 = 0

  1. Try factoring: (3x + 1)(4x + 3) = 12x2 + 13x + 3
  2. β†’ 3x + 1 = 0 or 4x + 3 = 0
  3. x = βˆ’1/3 or x = βˆ’3/4

(j) x2 + 3x + 2 = 0

  1. Numbers with product 2 and sum 3 β†’ 1 and 2
  2. (x + 1)(x + 2) = 0
  3. x = βˆ’1 or x = βˆ’2

(k) x2 βˆ’ 10x + 24 = 0

  1. Numbers with product 24 and sum βˆ’10 β†’ βˆ’6 and βˆ’4
  2. (x βˆ’ 6)(x βˆ’ 4) = 0
  3. x = 6 or x = 4

(l) 2x2 βˆ’ x βˆ’ 6 = 0

  1. Split βˆ’1x β†’ 3x βˆ’ 4x (since 2Β·(βˆ’6)=βˆ’12; 3 βˆ’ 4 = βˆ’1)
  2. (2x2 + 3x) + (βˆ’4x βˆ’ 6) = x(2x + 3) βˆ’ 2(2x + 3)
  3. (2x + 3)(x βˆ’ 2) = 0
  4. x = 2 or x = βˆ’3/2

(m) 3x + 2 = 9x2

  1. Rearrange: 9x2 βˆ’ 3x βˆ’ 2 = 0
  2. Factor: (3x βˆ’ 2)(3x + 1) = 0
  3. x = 2/3 or x = βˆ’1/3

(n) βˆ’3x2 + 11x βˆ’ 10 = 0

  1. Multiply by βˆ’1: 3x2 βˆ’ 11x + 10 = 0
  2. Factor: (3x βˆ’ 5)(x βˆ’ 2) = 0
  3. x = 5/3 or x = 2

(o) 4x2 = 25

  1. Bring to 0: 4x2 βˆ’ 25 = 0
  2. Use difference of squares: (2x βˆ’ 5)(2x + 5) = 0
  3. x = 5/2 or x = βˆ’5/2

(p) y2 βˆ’ 36 = 0

  1. Difference of squares: (y βˆ’ 6)(y + 6) = 0
  2. y = 6 or y = βˆ’6

(q) (x βˆ’ 8)2 = 36

  1. Square-root both sides (reason for Β±): x βˆ’ 8 = ±√36 = Β±6
  2. x = 8 + 6 = 14 or x = 8 βˆ’ 6 = 2

(r) 4x2 = 20x βˆ’ 25

  1. Rearrange: 4x2 βˆ’ 20x + 25 = 0
  2. Perfect square: (2x βˆ’ 5)2 = 0
  3. So 2x βˆ’ 5 = 0 β‡’ x = 5/2 (double root)

(s) 9y2 βˆ’ 6y + 1 = 0

  1. Perfect square: (3y βˆ’ 1)2 = 0
  2. 3y βˆ’ 1 = 0 β‡’ y = 1/3

(t) (x + 3)2 βˆ’ 49 = 0

  1. (x + 3)2 = 49
  2. Square-root both sides: x + 3 = Β±7
  3. x = 4 or x = βˆ’10
Why factorization works: We convert the equation to a product that equals 0 and then use the Zero-Product Property (if AB = 0, then A = 0 or B = 0). This is a fundamental property of real numbers, used implicitly since ancient mathematics.

2) Solve by completing the square

Idea: For x2 + bx, add (b/2)2 to form a perfect square (x + b/2)2. This approach was clearly taught by Al-Khwarizmi (~820 CE). We use Β± when taking square roots.

(a) x2 + 6x + 7 = 0

  1. Move constant: x2 + 6x = βˆ’7
  2. Add (6/2)2 = 9 both sides: x2 + 6x + 9 = 2
  3. Left is a square: (x + 3)2 = 2
  4. Take roots: x + 3 = ±√2 β‡’ x = βˆ’3 Β± √2

(b) x2 βˆ’ 11x + 1 = 0

  1. x2 βˆ’ 11x = βˆ’1
  2. Add (11/2)2 = 121/4: (x βˆ’ 11/2)2 = 121/4 βˆ’ 1 = 117/4
  3. x βˆ’ 11/2 = ±√(117/4) = Β±(√117)/2 = Β±(3√13)/2
  4. x = (11 ± 3√13)/2

(c) x2 βˆ’ 7x βˆ’ 7 = 0

  1. x2 βˆ’ 7x = 7
  2. Add (7/2)2 = 49/4: (x βˆ’ 7/2)2 = 7 + 49/4 = 77/4
  3. x βˆ’ 7/2 = ±√(77/4) = ±√77/2
  4. x = 7/2 ± √77/2

(d) 2x2 βˆ’ 10x + 7 = 0

  1. Divide by 2: x2 βˆ’ 5x + 7/2 = 0 β‡’ x2 βˆ’ 5x = βˆ’7/2
  2. Add (5/2)2 = 25/4: (x βˆ’ 5/2)2 = 25/4 βˆ’ 7/2 = 11/4
  3. x βˆ’ 5/2 = ±√(11/4) = ±√11/2
  4. x = 5/2 ± √11/2

(e) m2 + 5m = 1

  1. m2 + 5m βˆ’ 1 = 0 β‡’ m2 + 5m = 1
  2. Add (5/2)2 = 25/4: (m + 5/2)2 = 1 + 25/4 = 29/4
  3. m + 5/2 = ±√(29/4) = ±√29/2
  4. m = βˆ’5/2 Β± √29/2

(f) p2 βˆ’ 10p + 5 = 0

  1. p2 βˆ’ 10p = βˆ’5
  2. Add (10/2)2 = 25: (p βˆ’ 5)2 = 20
  3. p βˆ’ 5 = ±√20 = Β±2√5
  4. p = 5 ± 2√5

(g) 2b2 = 8b + 11

  1. 2b2 βˆ’ 8b βˆ’ 11 = 0 β‡’ divide by 2: b2 βˆ’ 4b βˆ’ 11/2 = 0
  2. b2 βˆ’ 4b = 11/2; add (4/2)2 = 4: (b βˆ’ 2)2 = 11/2 + 4 = 19/2
  3. b βˆ’ 2 = ±√(19/2) = Β±(√38)/2
  4. b = 2 ± √(19/2) (same as 2 ± √38/2)

(h) 3a2 βˆ’ 12a = 2

  1. 3a2 βˆ’ 12a βˆ’ 2 = 0 β‡’ divide by 3: a2 βˆ’ 4a βˆ’ 2/3 = 0
  2. a2 βˆ’ 4a = 2/3; add (4/2)2 = 4: (a βˆ’ 2)2 = 2/3 + 4 = 14/3
  3. a βˆ’ 2 = ±√(14/3)
  4. a = 2 ± √(14/3)

(i) 5n2 = 20n + 28

  1. 5n2 βˆ’ 20n βˆ’ 28 = 0 β‡’ divide by 5: n2 βˆ’ 4n βˆ’ 28/5 = 0
  2. n2 βˆ’ 4n = 28/5; add 4: (n βˆ’ 2)2 = 28/5 + 4 = 48/5
  3. n βˆ’ 2 = ±√(48/5) = Β±(4√15)/5
  4. n = 2 ± (4√15)/5

(j) c2 βˆ’ 8c + 13 = 0

  1. c2 βˆ’ 8c = βˆ’13
  2. Add (8/2)2 = 16: (c βˆ’ 4)2 = 3
  3. c βˆ’ 4 = ±√3
  4. c = 4 ± √3
Why completing the square works: it converts a quadratic into a perfect square so we can take square roots (using Β±). This method is historically tied to Al-Khwarizmi, and from it one can derive the modern quadratic formula.
Everyday link

Suppose you frame a 12 cm Γ— 9 cm photo with a border of width x. The visible area is quadratic in x. Solving for x (given a border area) uses exactly the same steps you learned above.

Math Questions – Full Answers

Mathematics Questions – Step-by-Step Answers

3) Use the quadratic formula to solve:

Quadratic formula: for axΒ² + bx + c = 0, x = (βˆ’b Β± √(bΒ² βˆ’ 4ac)) / (2a). (Derived from completing the square; classically traced to Al-Khwarizmi, ~820 CE.)

  1. 4xΒ² βˆ’ 4x + 1 = 0
    a=4, b=βˆ’4, c=1 β†’ Ξ”=bΒ²βˆ’4ac=16βˆ’16=0 β†’ βˆšΞ”=0.
    x=(βˆ’bΒ±0)/(2a)=(4)/8=1/2 (double root).
  2. 5xΒ² + 12x + 3 = 0
    a=5, b=12, c=3 β†’ Ξ”=144βˆ’60=84=4Β·21 β†’ βˆšΞ”=2√21.
    x=(βˆ’12 Β± 2√21)/10 = (βˆ’6 Β± √21)/5.
  3. (3xβˆ’2)(2xβˆ’5) = 5x(xβˆ’2)
    Expand LHS: 6xΒ² βˆ’19x + 10; RHS: 5xΒ² βˆ’ 10x.
    Bring to 0: xΒ² βˆ’ 9x + 10 = 0 β†’ factor: (xβˆ’5)(xβˆ’2)=0.
    x = 5 or 2.
  4. 3xΒ² = 4x + 4
    3xΒ² βˆ’ 4x βˆ’ 4 = 0; a=3,b=βˆ’4,c=βˆ’4 β†’ Ξ”=16+48=64 β†’ βˆšΞ”=8.
    x=(4 Β± 8)/6 β†’ x=2 or x=βˆ’2/3.
  5. 2xΒ² βˆ’ 5x + 2 = 0
    a=2,b=βˆ’5,c=2 β†’ Ξ”=25βˆ’16=9 β†’ βˆšΞ”=3.
    x=(5 Β± 3)/4 β†’ x=2 or x=1/2.
  6. 5xΒ² βˆ’ x βˆ’ 18 = 0
    a=5,b=βˆ’1,c=βˆ’18 β†’ Ξ”=1+360=361=19Β² β†’ βˆšΞ”=19.
    x=(1 Β± 19)/10 β†’ x=2 or x=βˆ’9/5.
4) Ages

Let son’s age be s, man’s age is 4s. In 4 years: (s+4)(4s+4)=520.
Expand β†’ 4sΒ² + 20s + 16 = 520 β†’ 4sΒ² + 20s βˆ’ 504 = 0 β†’ divide 4:
sΒ² + 5s βˆ’ 126 = 0 β†’ (s+14)(sβˆ’9)=0 β†’ s=9 (positive).
Son’s present age: 9 years.

5) Pencil prices

Let expensive price be p, cheap price pβˆ’40 (shillings). Budget 1800.
If buying expensive: number n with np=1800. Cheap gives 12 more: (n+12)(pβˆ’40)=1800.
Substitute n=1800/p β†’ (1800/p + 12)(pβˆ’40)=1800 β†’ simplify β†’ 12pΒ² βˆ’ 480p βˆ’ 72000 = 0.
Divide 12 β†’ pΒ² βˆ’ 40p βˆ’ 6000 = 0; Ξ”=1600+24000=25600=160Β².
p=(40 Β± 160)/2 β†’ p=100 (valid) or βˆ’60 (reject). Cheap price = 100βˆ’40=60.
Prices: Expensive 100/-, Cheaper 60/-. (Check: 1800/100=18, 1800/60=30 β†’ 12 more βœ“)

6) Consecutive numbers with squares summing to 145

Let numbers be n and n+1.
nΒ² + (n+1)Β² = 145 β†’ 2nΒ² + 2n + 1 = 145 β†’ 2nΒ² + 2n βˆ’ 144 = 0 β†’ divide 2:
nΒ² + n βˆ’ 72 = 0 β†’ (n+9)(nβˆ’8)=0 β†’ n=8 (or βˆ’9).
Consecutive numbers: 8 and 9. (Also βˆ’9 and βˆ’8 if negatives allowed.)

7) Picture frame width

Inner picture: 12 cm Γ— 9 cm (area 108). Border width = x all around.
Outer size: (12+2x) by (9+2x). Frame area = outer βˆ’ inner = 180.
(12+2x)(9+2x) βˆ’ 108 = 180 β†’ (12+2x)(9+2x) = 288.
Expand: 4xΒ² + 42x + 108 = 288 β†’ 4xΒ² + 42x βˆ’ 180 = 0 β†’ divide 2:
2xΒ² + 21x βˆ’ 90 = 0. Using quadratic formula: x = [βˆ’21 Β± √(21Β² βˆ’ 4Β·2Β·(βˆ’90))]/(4)
= [βˆ’21 Β± √(441 + 720)]/4 = [βˆ’21 Β± √1161]/4.
Positive solution: x = (βˆ’21 + √1161)/4 β‰ˆ 3.27 cm.

8) Find the number

Let the number be n. Condition: 6/n = (n+5)/6.
Cross-multiply: 36 = n(n+5) β†’ nΒ² + 5n βˆ’ 36 = 0 β†’ (n+9)(nβˆ’4)=0.
n = 4 or n = βˆ’9. (Both satisfy the statement.)

9) Whole number where twice its square is 11 more than 21 times the number

Let the number be x. Equation: 2xΒ² = 21x + 11 β†’ 2xΒ² βˆ’ 21x βˆ’ 11 = 0.
Ξ” = (βˆ’21)Β² βˆ’ 4Β·2Β·(βˆ’11) = 441 + 88 = 529 = 23Β².
x = [21 Β± 23]/4 β†’ x = 44/4 = 11 (or βˆ’2/4 = βˆ’1/2, not whole).
Answer: x = 11.

10) Rectangle from 56 cm wire; area 171 cmΒ²

Let sides be a and b (cm). Perimeter 2(a+b)=56 β†’ a+b=28. Area ab=171.
Put b=28βˆ’a β†’ a(28βˆ’a)=171 β†’ 28a βˆ’ aΒ² βˆ’ 171 = 0 β†’ aΒ² βˆ’ 28a + 171 = 0.
Ξ” = 28Β² βˆ’ 4Β·171 = 784 βˆ’ 684 = 100 β†’ βˆšΞ” = 10.
a = [28 Β± 10]/2 β†’ a = 19 or 9. Then b = 9 or 19.
Dimensions: 9 cm by 19 cm.

Reference Book: N/A

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