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State Board · Class 10 · Mathematics Part 1

Lesson 2: Quadratic Equations

5 concepts · 40 practice questions with answers · Free to read

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Quadratic equation: Introduction

From daily life

A farmer has a rectangular field of area 150 m2. Its length is 5 m more than its breadth x. Area = length × breadth gives x(x + 5) = 150, that is x2 + 5x − 150 = 0. This is a quadratic equation.

Simple definition

A quadratic equation (द्विघातीय समीकरण) in x has the general form ax2 + bx + c = 0, where a, b, c are real numbers and a ≠ 0.

Rule 1: Identifying a quadratic equation — It has one variable, all powers are whole numbers, and the highest power is 2. x2 − 9 = 0 is x2 + 0x − 9 = 0 with a = 1, b = 0, c = −9. (x + 3)(x − 4) = 0 expands to x2 − x − 12 = 0, which is quadratic. But x3 − 2x + 1 = 0 is not, because the highest power is 3.

  • Easy : Easy: In 2x2 − 5x + 1 = 0: a = 2, b = −5, c = 1.
  • Medium : Medium: (x + 1)(x + 2) = 0 becomes x2 + 3x + 2 = 0.
  • Exam-type : Exam: x(x + 5) = 150 gives x2 + 5x − 150 = 0 with a = 1, b = 5, c = −150.

Rule 2: Roots — The values of x that satisfy the equation are its roots (मुळे; solutions). If p(a) = 0, then a is a root and (x − a) is a factor of p(x). For x2 − 3x − 10: at x = 5, 25 − 15 − 10 = 0, so 5 is a root; at x = 1, 1 − 3 − 10 = −12, so 1 is not. If x = 2 is a root of x2 + kx − 10 = 0, then 4 + 2k − 10 = 0, so k = 3.

The parabola y = x squared minus 4x plus 3 crosses the x axis at 1 and 3, the roots−112345−2−112345xyy = x² − 4x + 3 and its rootsx = 1x = 3lowest point(x − 1)(x − 3) = 0

In the figure : In the figure the roots of a quadratic equation are where its graph meets the x axis.

  • Easy : Easy: x = 3 is a root of x2 − 9 = 0.
  • Medium : Medium: Is x = 4 a root of x2 − 3x − 4 = 0? 16 − 12 − 4 = 0, yes.
  • Exam-type : Exam: x = 3 is a root of 2x2 − kx + 3 = 0: 18 − 3k + 3 = 0, so k = 7.
Remember

a ≠ 0. The highest power is 2. A root makes the equation true.

Common mistake

Forgetting that a cannot be 0. If a = 0 the equation is linear.

Shortcut

Memory clue: "घात 2, a ≠ 0".

Spot the difference

EquationQuadratic?
x2 − 9 = 0Yes
(x + 3)(x − 4) = 0Yes
x3 − 2x + 1 = 0No

Practice questions

Easy
  1. 1What is the general form of a quadratic equation?
  2. 2What is the highest power in a quadratic equation?
  3. 3Is 5x2 − 3 = 0 quadratic?
Medium
  1. 4Is x3 + x = 2 quadratic?
  2. 5What is a root?
  3. 6Is x = 2 a root of x2 − 5x + 6 = 0?
Challenging
  1. 7x = −1 is a root of x2 + kx + 6 = 0. Find k.
  2. 8Write (2x − 1)(x + 3) = 5 in general form.
Show answers (try them yourself first)
  1. 1ax2 + bx + c = 0, a ≠ 0.
  2. 22.
  3. 3Yes, a = 5, b = 0, c = −3.
  4. 4No.
  5. 5A value of x that satisfies the equation.
  6. 64 − 10 + 6 = 0, yes.
  7. 71 − k + 6 = 0, k = 7.
  8. 82x2 + 5x − 3 = 5, so 2x2 + 5x − 8 = 0.

Methods of solving quadratic equation

From daily life

To split a 12 m rope into two pieces whose product of lengths is 35, you can try pairs. But for harder problems we need dependable methods: factorisation, completing the square and the formula.

Simple definition

Three methods solve a quadratic equation: factorisation, completing the square and the formula x = −b ± √(b2 − 4ac)2a.

Rule 1: Factorisation — If the product of two numbers is 0, at least one is 0. x2 − 7x + 10 = (x − 5)(x − 2) = 0 gives x = 5 or 2. 2x2 + 7x + 3 = (2x + 1)(x + 3) = 0 gives x = −12 or −3. 4y2 = 12y gives 4y(y − 3) = 0, so y = 0 or 3. x2 = 5 gives x = √5 or −√5.

  • Easy : Easy: x2 − 5x + 6 = (x − 2)(x − 3), roots 2 and 3.
  • Medium : Medium: x2 + x − 12 = (x + 4)(x − 3), roots −4 and 3.
  • Exam-type : Exam: 3x2 − 2x = 0 gives x(3x − 2) = 0, so x = 0 or 23.

Rule 2: Completing the square — x2 + 6x + 1 = 0: add and subtract 9 to get (x + 3)2 − 8 = 0, so x + 3 = ±2√2 and x = −3 ± 2√2. x2 − 8x − 20 = 0: (x − 4)2 = 36, so x = 4 ± 6, that is 10 or −2. If the coefficient of x2 is not 1, divide first: 2x2 + 4x − 6 = 0 becomes x2 + 2x − 3 = 0, (x + 1)2 = 4, x = 1 or −3.

A big square of side x plus 3 made of an x by x square, two 3 by x rectangles and a missing 3 by 3 cornerCompleting the square: x² + 6xx²3x3x9x3x3x² + 6x = blue + orangeadd the green square 9x² + 6x + 9 = (x + 3)²so x² + 6x = (x + 3)² − 9The missing corner has area (half of 6)² = 9

In the figure : In the figure completing the square drawn as areas: the missing corner square (9) makes (x + 3)2.

  • Easy : Easy: x2 + 2x + 1 = (x + 1)2, so add 1 to complete x2 + 2x.
  • Medium : Medium: x2 + 10x needs 25 to become (x + 5)2.
  • Exam-type : Exam: x2 − 4x − 1 = 0: (x − 2)2 = 5, so x = 2 ± √5.

Rule 3: The formula — For ax2 + bx + c = 0, x = −b ± √(b2 − 4ac)2a. x2 − 5x + 6 = 0: b2 − 4ac = 25 − 24 = 1, x = 5 ± 12 = 3 or 2. 2x2 + 3x − 2 = 0: 9 + 16 = 25, x = −3 ± 54 = 12 or −2. x2 + 4x + 1 = 0: 16 − 4 = 12, x = −4 ± 2√32 = −2 ± √3. x2 − x − 6 = 0 gives 3 and −2 by all three methods.

  • Easy : Easy: In x2 − 7x + 12 = 0: a = 1, b = −7, c = 12.
  • Medium : Medium: Its b2 − 4ac = 49 − 48 = 1, so x = 7 ± 12 = 4 or 3.
  • Exam-type : Exam: x2 + x + 1 = 0 has b2 − 4ac = 1 − 4 = −3, so no real roots.
Remember

Bring the equation to ax2 + bx + c = 0 first. Then pick a method.

Common mistake

Dividing both sides by x. This loses the root x = 0. Factor instead.

Shortcut

Memory clue: "अवयव → वर्ग → सूत्र".

Spot the difference

MethodBest when
FactorisationFactors are easy to see
Completing the squareLearning the formula's origin
FormulaAlways works

Practice questions

Easy
  1. 1Name three methods of solving a quadratic equation.
  2. 2Write the quadratic formula.
  3. 3Solve x2 − 9x + 20 = 0.
Medium
  1. 4Solve x2 = 49.
  2. 5What must be added to x2 + 12x to complete the square?
  3. 6Solve 2x2 − 5x − 3 = 0.
Challenging
  1. 7Solve x2 + 6x − 7 = 0 by completing the square.
  2. 8Solve x2 − 6x + 4 = 0 by the formula.
Show answers (try them yourself first)
  1. 1Factorisation, completing the square, formula.
  2. 2x = −b ± √(b2 − 4ac)2a.
  3. 3(x − 4)(x − 5) = 0, x = 4 or 5.
  4. 4x = 7 or −7.
  5. 536.
  6. 6(2x + 1)(x − 3) = 0, x = −12 or 3.
  7. 7(x + 3)2 = 16, x = 1 or −7.
  8. 8b2 − 4ac = 20, x = 6 ± 2√52 = 3 ± √5.

Nature of roots of quadratic equation

From daily life

You cannot always know in advance whether a quadratic equation has answers that are real numbers. A single number, b2 − 4ac, tells this before solving, like a weather forecast.

Simple definition

The discriminant (विवेचक) is Δ = b2 − 4ac. It decides the nature of the roots.

Rule 1: Nature of roots — If Δ > 0, the roots are real and unequal. If Δ = 0, the roots are real and equal. If Δ < 0, the roots are not real. x2 + x − 6 = 0: Δ = 1 + 24 = 25 > 0, real unequal. x2 + 6x + 9 = 0: Δ = 36 − 36 = 0, real equal. 2x2 − 3x + 5 = 0: Δ = 9 − 40 = −31 < 0, not real.

Three parabolas showing the discriminant positive, zero and negativeΔ > 0two different real rootsΔ = 0two equal rootsΔ < 0no real roots

In the figure : In the figure the discriminant tells how many times the graph meets the x axis.

Live figure

Nature of roots of quadratic equation

In the app you can move this figure with your finger.

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  • Easy : Easy: x2 − 4x + 4 = 0 has Δ = 16 − 16 = 0.
  • Medium : Medium: x2 + 3x + 5 = 0 has Δ = 9 − 20 = −11, so no real roots.
  • Exam-type : Exam: 3x2 − 5x + 1 = 0 has Δ = 25 − 12 = 13 > 0, real and unequal.

Rule 2: Finding an unknown from the nature of roots — For equal roots put Δ = 0. x2 + kx + 9 = 0: k2 − 36 = 0, so k = ±6. 2x2 + kx + 8 = 0: k2 − 64 = 0, so k = ±8. kx2 − 12x + 9 = 0: 144 − 36k = 0, so k = 4. x2 + kx + k = 0: k2 − 4k = 0, so k = 0 or 4.

  • Easy : Easy: x2 + kx + 4 = 0 has equal roots when k2 = 16, k = ±4.
  • Medium : Medium: kx2 − 4x + 1 = 0: 16 − 4k = 0, k = 4.
  • Exam-type : Exam: For real roots Δ ≥ 0: x2 + 4x + k = 0 needs 16 − 4k ≥ 0, so k ≤ 4.
Remember

Find Δ = b2 − 4ac first. Δ < 0 means no real roots.

Common mistake

Taking the sign of b wrongly. In Δ the b is squared, so b2 is always positive.

Shortcut

Memory clue: "Δ धन = दोन, शून्य = एक, ऋण = शून्य मुळे".

Spot the difference

ΔRootsGraph
Δ > 0Real, unequalCuts the x-axis twice
Δ = 0Real, equalTouches the x-axis
Δ < 0Not realDoes not meet the x-axis

Practice questions

Easy
  1. 1What is the discriminant?
  2. 2What does Δ = 0 mean?
  3. 3What does Δ < 0 mean?
Medium
  1. 4Find Δ for x2 + 5x + 6.
  2. 5Find Δ for 2x2 − 4x + 5.
  3. 6For what k does x2 + kx + 16 = 0 have equal roots?
Challenging
  1. 7Does 5x2 + 2x − 3 = 0 have real roots?
  2. 8Why can a quadratic graph touch the x-axis at one point only when Δ = 0?
Show answers (try them yourself first)
  1. 1Δ = b2 − 4ac.
  2. 2Real and equal roots.
  3. 3The roots are not real.
  4. 425 − 24 = 1.
  5. 516 − 40 = −24, roots not real.
  6. 6k2 = 64, k = ±8.
  7. 7Δ = 4 + 60 = 64 > 0, yes, unequal.
  8. 8Both roots coincide, giving one common point.

Relation between roots and coefficients

From daily life

If you know two numbers, you can find their sum and product easily. Surprisingly, you can often find the sum and product of the roots without solving the equation at all, just by reading the coefficients.

Simple definition

If α and β are the roots of ax2 + bx + c = 0, then α + β = −ba and α × β = ca.

Rule 1: Sum and product — For 2x2 + 8x + 3 = 0: α + β = −82 = −4 and αβ = 32. For x2 − 9x + 14 = 0: sum 9 and product 14 (the roots are 2 and 7). Useful: α2 + β2 = (α + β)2 − 2αβ. For x2 + 4x − 3 = 0: 16 + 6 = 22. If the roots of x2 − 11x + k = 0 differ by 3: α + β = 11, α − β = 3, so α = 7, β = 4 and k = 28.

  • Easy : Easy: x2 − 6x + 5 = 0: sum 6, product 5.
  • Medium : Medium: 3x2 + 6x − 9 = 0: sum −2, product −3.
  • Exam-type : Exam: x2 + 4x − 3 = 0: α2 + β2 = (−4)2 − 2(−3) = 22.

Rule 2: Forming an equation from its roots — The equation with roots α and β is x2 − (α + β)x + αβ = 0. Roots 4 and −1: sum 3, product −4, so x2 − 3x − 4 = 0. Roots 2 and 6: x2 − 8x + 12 = 0. Roots 3 + √2 and 3 − √2: sum 6, product 9 − 2 = 7, so x2 − 6x + 7 = 0. If the sum of the roots of x2 − 3kx + (k + 3) = 0 is twice their product: 3k = 2(k + 3), so k = 6.

Roots 4 and 6 give sum 10, product 24 and the equation x squared minus 10x plus 24Building the equation from its rootsRoots 4 and 6Sum = 10Product = 24x² − 10x + 24 = 0

In the figure : In the figure a quadratic equation from its roots: x2 − (sum)x + (product) = 0.

  • Easy : Easy: Roots 1 and 5 give x2 − 6x + 5 = 0.
  • Medium : Medium: Roots 0 and 4 give x2 − 4x = 0.
  • Exam-type : Exam: Sum 6 and product 5 give x2 − 6x + 5 = 0, whose roots are 1 and 5.
Remember

Sum = −b ÷ a. Product = c ÷ a. Equation: x2 − (sum)x + product.

Common mistake

Forgetting the minus sign in the sum. Sum is −ba.

Shortcut

Memory clue: "बेरीज = −b/a; गुणाकार = c/a".

Spot the difference

QuantityValue
α + β−ba
α × βca
Equationx2 − (α + β)x + αβ = 0

Practice questions

Easy
  1. 1State the sum of the roots.
  2. 2State the product of the roots.
  3. 3x2 − 7x + 12 = 0: sum and product?
Medium
  1. 42x2 + 5x − 3 = 0: sum and product?
  2. 5Make the equation with roots 2 and −5.
  3. 6Make the equation with roots 1 + √3 and 1 − √3.
Challenging
  1. 7For x2 − 5x + 2 = 0, find α2 + β2.
  2. 8Roots of x2 − 9x + k = 0 differ by 5. Find k.
Show answers (try them yourself first)
  1. 1−ba.
  2. 2ca.
  3. 37 and 12.
  4. 4−52 and −32.
  5. 5x2 + 3x − 10 = 0.
  6. 6Sum 2, product −2: x2 − 2x − 2 = 0.
  7. 725 − 4 = 21.
  8. 8α = 7, β = 2, k = 14.

Applications of quadratic equations

From daily life

A fencing problem, a speed problem and a number puzzle can all hide a quadratic equation. After solving it, the story decides which root to keep: a length, speed or age can never be negative.

Simple definition

To solve a word problem: assume the unknown as x, form the quadratic equation, solve it, and reject the root that does not fit the situation.

Rule 1: Geometry and numbers — A right triangle has hypotenuse 13 cm and one leg 7 cm longer than the other. x2 + (x + 7)2 = 169 gives x2 + 7x − 60 = 0, (x + 12)(x − 5) = 0. Reject x = −12; x = 5, so the legs are 5 cm and 12 cm. The product of two consecutive natural numbers is 156: x(x + 1) = 156, x2 + x − 156 = 0, (x + 13)(x − 12) = 0, so 12 and 13. A garden's length is 3 m more than its breadth and its area is 180 m2: x2 + 3x − 180 = 0, (x + 15)(x − 12) = 0, so breadth 12 m and length 15 m.

Four steps for a quadratic word problemSolving a word problemTake x asthe unknownForm theequationSolve thequadraticReject theunsuitable root

In the figure : In the figure steps for applications; a negative length, age or speed is always rejected.

  • Easy : Easy: Two consecutive numbers with product 20: x(x + 1) = 20 gives 4 and 5.
  • Medium : Medium: Check the garden: 12 × 15 = 180.
  • Exam-type : Exam: Check the triangle: 52 + 122 = 25 + 144 = 169 = 132.

Rule 2: Speed and time — A bus covers 120 km. If its speed were 10 km/h more, it would take 1 hour less. With speed x: 120x − 120(x + 10) = 1. Multiply by x(x + 10): 120(x + 10) − 120x = x(x + 10), so x2 + 10x − 1200 = 0, (x + 40)(x − 30) = 0. Reject x = −40; the speed is 30 km/h. Check: 120 ÷ 30 = 4 h and 120 ÷ 40 = 3 h.

  • Easy : Easy: Time = distance ÷ speed.
  • Medium : Medium: At 30 km/h, 120 km takes 4 hours.
  • Exam-type : Exam: Why reject −40? A speed cannot be negative.
Remember

Form the equation, solve it, reject unsuitable roots, then check in the story.

Common mistake

Giving both roots as answers. Negative lengths, ages and speeds must be dropped.

Shortcut

Memory clue: "समीकरण → सोडवा → नाकारा → तपासा".

Spot the difference

ProblemEquationAnswer
Right trianglex2 + 7x − 60 = 05 cm, 12 cm
Consecutive numbersx2 + x − 156 = 012, 13
Bus speedx2 + 10x − 1200 = 030 km/h

Practice questions

Easy
  1. 1What is the first step in a word problem?
  2. 2Why do we reject some roots?
  3. 3Two consecutive natural numbers have product 30. Find them.
Medium
  1. 4A rectangle's length is 2 m more than its breadth and its area is 24 m2. Find the sides.
  2. 5The sum of a number and its square is 42. Find the positive number.
  3. 6A right triangle has legs x and x + 1 and hypotenuse 5. Find x.
Challenging
  1. 7A train covers 180 km. At 15 km/h more speed it takes 1 hour less. Find the original speed.
  2. 8Why must we check the answer in the story?
Show answers (try them yourself first)
  1. 1Assume the unknown as x.
  2. 2They may not fit the situation (negative length, age, speed).
  3. 3x2 + x − 30 = 0, x = 5; numbers 5 and 6.
  4. 4x2 + 2x − 24 = 0, x = 4; 4 m and 6 m.
  5. 5x2 + x − 42 = 0, (x + 7)(x − 6) = 0; x = 6.
  6. 62x2 + 2x − 24 = 0, x = 3 (reject −4); legs 3 and 4.
  7. 7x2 + 15x − 2700 = 0, (x + 60)(x − 45) = 0; 45 km/h.
  8. 8To be sure it satisfies the given conditions.

Quick revision

  • Quadratic equation: ax² + bx + c = 0 with a ≠ 0; roots by factorisation, completing the square or the formula x = −b ± √(b² − 4ac)2a.
  • Δ = b² − 4ac: Δ = 0 real and equal roots, Δ > 0 real and unequal roots, Δ < 0 roots not real.
  • α + β = −ba, α × β = ca, and the equation with roots α and β is x² − (α + β)x + αβ = 0.