Key Concepts & Formulae at a Glance
- Standard form:ax²+bx+c=0, where a, b, c are real numbers and a≠0.
- Root of a quadratic equation:α is a root if aα²+bα+c=0. Roots of the equation = zeroes of the corresponding polynomial.
- Solving by factorisation:Split the middle term so ax²+bx+c factors into two linear factors; set each factor to zero.
- Quadratic formula:x = (−b ± √(b²−4ac)) / 2a, provided b²−4ac ≥ 0.
- Discriminant:D = b²−4ac determines the nature of the roots.
- Nature of roots:D>0: two distinct real roots. D=0: two equal real roots. D<0: no real roots.
Exercise 4.1
2 Questions
1Check whether the following are quadratic equations:
(i) (x+1)² = 2(x−3)
(ii) x²−2x = (−2)(3−x)
(iii) (x−2)(x+1) = (x−1)(x+3)
(iv) (x−3)(2x+1) = x(x+5)
(v) (2x−1)(x−3) = (x+5)(x−1)
(vi) x²+3x+1 = (x−2)²
(vii) (x+2)³ = 2x(x²−1)
(viii) x³−4x²−x+1 = (x−2)³
(i) (x+1)² = 2(x−3)
LHS = x²+2x+1. Equation becomes x²+2x+1 = 2x−6
x²+2x+1−2x+6 = 0 → x²+7 = 0 — of the form ax²+bx+c=0
x²+2x+1−2x+6 = 0 → x²+7 = 0 — of the form ax²+bx+c=0
Quadratic equation.
(ii) x²−2x = (−2)(3−x)
RHS = −6+2x. Equation becomes x²−2x = −6+2x
x²−2x−2x+6 = 0 → x²−4x+6 = 0
x²−2x−2x+6 = 0 → x²−4x+6 = 0
Quadratic equation.
(iii) (x−2)(x+1) = (x−1)(x+3)
LHS = x²−x−2. RHS = x²+2x−3
x²−x−2 = x²+2x−3 → −3x+1 = 0 — the x² terms cancel, leaving a linear equation
x²−x−2 = x²+2x−3 → −3x+1 = 0 — the x² terms cancel, leaving a linear equation
Not a quadratic equation.
(iv) (x−3)(2x+1) = x(x+5)
LHS = 2x²−5x−3. RHS = x²+5x
2x²−5x−3 = x²+5x → x²−10x−3 = 0
2x²−5x−3 = x²+5x → x²−10x−3 = 0
Quadratic equation.
(v) (2x−1)(x−3) = (x+5)(x−1)
LHS = 2x²−7x+3. RHS = x²+4x−5
2x²−7x+3 = x²+4x−5 → x²−11x+8 = 0
2x²−7x+3 = x²+4x−5 → x²−11x+8 = 0
Quadratic equation.
(vi) x²+3x+1 = (x−2)²
RHS = x²−4x+4
x²+3x+1 = x²−4x+4 → 7x−3 = 0 — the x² terms cancel, leaving a linear equation
x²+3x+1 = x²−4x+4 → 7x−3 = 0 — the x² terms cancel, leaving a linear equation
Not a quadratic equation.
(vii) (x+2)³ = 2x(x²−1)
LHS = x³+6x²+12x+8. RHS = 2x³−2x
x³+6x²+12x+8 = 2x³−2x → −x³+6x²+14x+8 = 0, i.e. x³−6x²−14x−8 = 0 — the x³ term survives
x³+6x²+12x+8 = 2x³−2x → −x³+6x²+14x+8 = 0, i.e. x³−6x²−14x−8 = 0 — the x³ term survives
Not a quadratic equation (it's cubic).
(viii) x³−4x²−x+1 = (x−2)³
RHS = x³−6x²+12x−8
x³−4x²−x+1 = x³−6x²+12x−8 → 2x²−13x+9 = 0 — the x³ terms cancel, a genuine quadratic remains
x³−4x²−x+1 = x³−6x²+12x−8 → 2x²−13x+9 = 0 — the x³ terms cancel, a genuine quadratic remains
Quadratic equation.
2Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan's mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan's present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 kmh less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
(i) Rectangular plot
Let the breadth = x m, so the length = (2x + 1) m.
Area = length × breadth: x(2x + 1) = 528
Area = length × breadth: x(2x + 1) = 528
2x² + x − 528 = 0
(ii) Consecutive positive integers
Let the integers be x and x + 1.
x(x + 1) = 306
x(x + 1) = 306
x² + x − 306 = 0
(iii) Rohan and his mother
Let Rohan's present age = x years, so his mother's present age = (x + 26) years.
3 years from now: Rohan = x+3, mother = x+29
(x + 3)(x + 29) = 360
3 years from now: Rohan = x+3, mother = x+29
(x + 3)(x + 29) = 360
x² + 32x − 273 = 0
(iv) Train speed
Let the usual speed = x kmh. Time taken = 480x hours.
At speed (x−8) kmh, time taken = 480x−8 hours, which is 3 hours more:
480x−8 − 480x = 3
Multiplying through by x(x−8): 480x − 480(x−8) = 3x(x−8)
3840 = 3x² − 24x
3x² − 24x − 3840 = 0, dividing by 3:
At speed (x−8) kmh, time taken = 480x−8 hours, which is 3 hours more:
480x−8 − 480x = 3
Multiplying through by x(x−8): 480x − 480(x−8) = 3x(x−8)
3840 = 3x² − 24x
3x² − 24x − 3840 = 0, dividing by 3:
x² − 8x − 1280 = 0
Exercise 4.2
6 Questions
1Find the roots of the following quadratic equations by factorisation:
(i) x²−3x−10=0 (ii) 2x²+x−6=0
(iii) √2x²+7x+5√2=0 (iv) 2x²−x+18=0
(v) 100x²−20x+1=0
(i) x²−3x−10=0
Splitting the middle term: −3x = 2x−5x (since 2×−5=−10)
x²+2x−5x−10=0 → x(x+2)−5(x+2)=0 → (x+2)(x−5)=0
x²+2x−5x−10=0 → x(x+2)−5(x+2)=0 → (x+2)(x−5)=0
x = −2 or x = 5
(ii) 2x²+x−6=0
Splitting the middle term: x = 4x−3x (since 4×−3=−12=2×−6)
2x²+4x−3x−6=0 → 2x(x+2)−3(x+2)=0 → (x+2)(2x−3)=0
2x²+4x−3x−6=0 → 2x(x+2)−3(x+2)=0 → (x+2)(2x−3)=0
x = −2 or x = 32
(iii) √2x²+7x+5√2=0
Splitting the middle term: 7x = 2x+5x (since 2×5√2×√2=... verify: 2 and 5√2 need product = √2×5√2=10)
√2x²+2x+5x+5√2=0 → √2x(x+√2)+5(x+√2)=0 → (x+√2)(√2x+5)=0
√2x²+2x+5x+5√2=0 → √2x(x+√2)+5(x+√2)=0 → (x+√2)(√2x+5)=0
x = −√2 or x = −5√2 = −5√22
(iv) 2x²−x+18=0
Multiplying through by 8: 16x²−8x+1=0
This factors as a perfect square: (4x−1)²=0
This factors as a perfect square: (4x−1)²=0
x = 14 (repeated root)
(v) 100x²−20x+1=0
This factors as a perfect square: (10x−1)²=0
x = 110 (repeated root)
2Solve the problems given in Example 1 (marbles, and toys).
Marbles: x²−45x+324=0
Splitting the middle term: −45x = −36x−9x (since −36×−9=324)
x²−36x−9x+324=0 → x(x−36)−9(x−36)=0 → (x−36)(x−9)=0
x = 36 or x = 9. Both are valid: John and Jivanti each need at least 5 marbles to lose, and both 36 and 9 satisfy that (since 45−36=9≥5 and 45−9=36≥5).
x²−36x−9x+324=0 → x(x−36)−9(x−36)=0 → (x−36)(x−9)=0
x = 36 or x = 9. Both are valid: John and Jivanti each need at least 5 marbles to lose, and both 36 and 9 satisfy that (since 45−36=9≥5 and 45−9=36≥5).
John had 36 marbles and Jivanti had 9, or John had 9 and Jivanti had 36.
Toys: x²−55x+750=0
Splitting the middle term: −55x = −25x−30x (since −25×−30=750)
x²−25x−30x+750=0 → x(x−25)−30(x−25)=0 → (x−25)(x−30)=0
x = 25 or x = 30. Both are valid since the cost per toy (55−x) stays positive in each case.
x²−25x−30x+750=0 → x(x−25)−30(x−25)=0 → (x−25)(x−30)=0
x = 25 or x = 30. Both are valid since the cost per toy (55−x) stays positive in each case.
The number of toys produced was 25 or 30.
3Find two numbers whose sum is 27 and product is 182.
Let one number = x, so the other = 27 − x.
x(27 − x) = 182
27x − x² = 182 → x² − 27x + 182 = 0
Splitting the middle term: −27x = −13x−14x (since −13×−14=182)
x(x−13)−14(x−13)=0 → (x−13)(x−14)=0
x(27 − x) = 182
27x − x² = 182 → x² − 27x + 182 = 0
Splitting the middle term: −27x = −13x−14x (since −13×−14=182)
x(x−13)−14(x−13)=0 → (x−13)(x−14)=0
The numbers are 13 and 14.
4Find two consecutive positive integers, sum of whose squares is 365.
Let the integers be x and x + 1.
x² + (x+1)² = 365
x² + x² + 2x + 1 = 365 → 2x² + 2x − 364 = 0 → x² + x − 182 = 0
Splitting the middle term: x = 14x−13x (since 14×−13=−182)
x(x+14)−13(x+14)=0 → (x+14)(x−13)=0
x = −14 (rejected, must be positive) or x = 13
x² + (x+1)² = 365
x² + x² + 2x + 1 = 365 → 2x² + 2x − 364 = 0 → x² + x − 182 = 0
Splitting the middle term: x = 14x−13x (since 14×−13=−182)
x(x+14)−13(x+14)=0 → (x+14)(x−13)=0
x = −14 (rejected, must be positive) or x = 13
The integers are 13 and 14.
5The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Let the base = x cm, so the altitude = (x−7) cm.
By the Pythagoras theorem: base² + altitude² = hypotenuse²
x² + (x−7)² = 13²
x² + x² − 14x + 49 = 169 → 2x² − 14x − 120 = 0 → x² − 7x − 60 = 0
Splitting the middle term: −7x = −12x+5x (since −12×5=−60)
x(x−12)+5(x−12)=0 → (x−12)(x+5)=0
x = −5 (rejected, base can't be negative) or x = 12
By the Pythagoras theorem: base² + altitude² = hypotenuse²
x² + (x−7)² = 13²
x² + x² − 14x + 49 = 169 → 2x² − 14x − 120 = 0 → x² − 7x − 60 = 0
Splitting the middle term: −7x = −12x+5x (since −12×5=−60)
x(x−12)+5(x−12)=0 → (x−12)(x+5)=0
x = −5 (rejected, base can't be negative) or x = 12
Right triangle: base 12 cm, altitude 5 cm, hypotenuse 13 cm
Base = 12 cm, Altitude = 5 cm (a 5–12–13 right triangle).
6A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹90, find the number of articles produced and the cost of each article.
Let the number of articles produced = x.
Cost per article = (2x + 3) rupees.
Total cost: x(2x + 3) = 90
2x² + 3x − 90 = 0
Splitting the middle term: 3x = 15x−12x (since 15×−12=−180=2×−90)
2x²+15x−12x−90=0 → x(2x+15)−6(2x+15)=0 → (2x+15)(x−6)=0
x = −152 (rejected, count of articles can't be negative) or x = 6
Cost per article = 2(6) + 3 = 15
Cost per article = (2x + 3) rupees.
Total cost: x(2x + 3) = 90
2x² + 3x − 90 = 0
Splitting the middle term: 3x = 15x−12x (since 15×−12=−180=2×−90)
2x²+15x−12x−90=0 → x(2x+15)−6(2x+15)=0 → (2x+15)(x−6)=0
x = −152 (rejected, count of articles can't be negative) or x = 6
Cost per article = 2(6) + 3 = 15
6 articles were produced, at a cost of ₹15 each.
Exercise 4.3
5 Questions
1Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2x²−3x+5=0 (ii) 3x²−4√3x+4=0 (iii) 2x²−6x+3=0
(i) 2x²−3x+5=0
Here a=2, b=−3, c=5.
Discriminant D = b²−4ac = (−3)²−4(2)(5) = 9−40 = −31 < 0
Discriminant D = b²−4ac = (−3)²−4(2)(5) = 9−40 = −31 < 0
D < 0, so no real roots.
(ii) 3x²−4√3x+4=0
Here a=3, b=−4√3, c=4.
Discriminant D = (−4√3)²−4(3)(4) = 48−48 = 0
Since D=0, the roots are both −b2a = 4√36 = 2√33
Discriminant D = (−4√3)²−4(3)(4) = 48−48 = 0
Since D=0, the roots are both −b2a = 4√36 = 2√33
D = 0, so two equal real roots: x = 2√33, 2√33
(iii) 2x²−6x+3=0
Here a=2, b=−6, c=3.
Discriminant D = (−6)²−4(2)(3) = 36−24 = 12 > 0
x = −b ± √D2a = 6 ± √124 = 6 ± 2√34 = 3 ± √32
Discriminant D = (−6)²−4(2)(3) = 36−24 = 12 > 0
x = −b ± √D2a = 6 ± √124 = 6 ± 2√34 = 3 ± √32
D > 0, so two distinct real roots: x = 3+√32 and x = 3−√32
2Find the values of k for each of the following quadratic equations, so that they have two equal roots:
(i) 2x²+kx+3=0 (ii) kx(x−2)+6=0
(i) 2x²+kx+3=0
Here a=2, b=k, c=3. For equal roots, D=0:
k²−4(2)(3) = 0 → k² = 24 → k = ±√24 = ±2√6
k²−4(2)(3) = 0 → k² = 24 → k = ±√24 = ±2√6
k = 2√6 or k = −2√6
(ii) kx(x−2)+6=0
Expanding: kx²−2kx+6=0. Here a=k, b=−2k, c=6 (need k≠0 for this to be quadratic).
For equal roots, D=0: (−2k)²−4(k)(6)=0
4k²−24k=0 → 4k(k−6)=0 → k=0 or k=6
Since k=0 would make the equation linear (not quadratic), it's rejected.
For equal roots, D=0: (−2k)²−4(k)(6)=0
4k²−24k=0 → 4k(k−6)=0 → k=0 or k=6
Since k=0 would make the equation linear (not quadratic), it's rejected.
k = 6
3Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.
Let the breadth = x m, so the length = 2x m.
Area: x × 2x = 800 → 2x² = 800 → x² = 400
Discriminant of x²−400=0: D = 0²−4(1)(−400) = 1600 > 0 — real roots exist
x = √400 = 20 (rejecting x=−20, since breadth can't be negative)
Length = 2x = 40
Area: x × 2x = 800 → 2x² = 800 → x² = 400
Discriminant of x²−400=0: D = 0²−4(1)(−400) = 1600 > 0 — real roots exist
x = √400 = 20 (rejecting x=−20, since breadth can't be negative)
Length = 2x = 40
Yes, it's possible. Breadth = 20 m, Length = 40 m.
4Is the following situation possible? If so, determine their present ages.
The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Let one friend's present age = x years, so the other's = (20−x) years.
Four years ago: ages were (x−4) and (16−x).
(x−4)(16−x) = 48
16x−x²−64+4x = 48 → −x²+20x−64=48 → −x²+20x−112=0 → x²−20x+112=0
Discriminant: D = (−20)²−4(1)(112) = 400−448 = −48 < 0
Four years ago: ages were (x−4) and (16−x).
(x−4)(16−x) = 48
16x−x²−64+4x = 48 → −x²+20x−64=48 → −x²+20x−112=0 → x²−20x+112=0
Discriminant: D = (−20)²−4(1)(112) = 400−448 = −48 < 0
D < 0, so no real roots exist. This situation is not possible.
5Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.
Half the perimeter = length + width = 40 m. Let width = x m, so length = (40−x) m.
Area: x(40−x) = 400 → 40x−x²=400 → x²−40x+400=0
Discriminant: D = (−40)²−4(1)(400) = 1600−1600 = 0 — two equal real roots
x = −b2a = 402 = 20. So width = 20, length = 40−20 = 20
Area: x(40−x) = 400 → 40x−x²=400 → x²−40x+400=0
Discriminant: D = (−40)²−4(1)(400) = 1600−1600 = 0 — two equal real roots
x = −b2a = 402 = 20. So width = 20, length = 40−20 = 20
Yes, it's possible — both length and width are 20 m (the park is a square).
Frequently Asked Questions
Expand and simplify both sides fully, then bring everything to one side. If the highest power of x that survives is exactly 2 (the x² term doesn't cancel out), it's quadratic — even if the original expression looked cubic, like (x+2)³ = x³ − 4, where the x³ terms cancel and leave a genuine quadratic.
For ax² + bx + c = 0, the discriminant is b² − 4ac. If it's positive, the equation has two distinct real roots. If it's zero, the equation has two equal real roots. If it's negative, the equation has no real roots.
Split the middle term (bx) into two terms whose coefficients multiply to give a×c and add to give b, factor the resulting four-term expression by grouping into two linear factors, then set each factor equal to zero to get the two roots.
Check each root against the real-world constraints of the problem — lengths, ages, speeds, and counts of things can't be negative. Discard any root that would make a physical quantity negative or otherwise impossible, and keep the one that fits.
Yes — a printable Formula Card covering all 7 units, including Quadratic Equations, is available for ₹99.
Continue to the Next Chapter
Chapter 5 (Arithmetic Progressions) introduces sequences that build on the algebra from this chapter.