Chapter 4: Quadratic Equations
Quadratic equations
ax² + bx + c = 0 (a ≠ 0) is the standard form of a quadratic equation. Sometimes an equation must be simplified first. For example, (x + 2)³ = x³ − 4 simplifies to x² + 2x + 2 = 0, so it is quadratic, while x(x + 1) + 8 = (x + 2)(x − 2) simplifies to x + 12 = 0, which is not quadratic.
Finding roots by factorisation
A real number α is a root if aα² + bα + c = 0. The zeroes of the quadratic polynomial and the roots of the equation are the same, and there are at most two roots. Split the middle term to get two linear factors and equate each factor to zero.
Nature of roots and the quadratic formula
The discriminant is D = b² − 4ac. D > 0: two distinct real roots. D = 0: two equal real roots. D < 0: no real roots. The quadratic formula: x = (−b ± √(b² − 4ac)) / 2a, when b² − 4ac ≥ 0.
Word problems
Take the unknown as x, form the equation, solve it, and finally see which value is possible in the situation. In the prayer hall problem the roots for the breadth were 12 and −12.5, but a length or breadth cannot be negative, so the breadth is 12 m and the length is 25 m.
Important questions
Q1. What is a quadratic equation? Write its standard form.
Equating a polynomial of degree 2 in x to zero gives a quadratic equation. Standard form: ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0.
Q2. Check whether (x − 2)² + 1 = 2x − 3 is a quadratic equation.
The left side is x² − 4x + 5. So x² − 4x + 5 = 2x − 3, that is x² − 6x + 8 = 0. It is of the form ax² + bx + c = 0, so it is a quadratic equation.
Q3. Check whether x(x + 1) + 8 = (x + 2)(x − 2) is a quadratic equation.
x² + x + 8 = x² − 4, that is x + 12 = 0. It is not a quadratic equation, because the x² term cancels out.
Q4. Check whether (x + 2)³ = x³ − 4 is a quadratic equation.
x³ + 6x² + 12x + 8 = x³ − 4, that is 6x² + 12x + 12 = 0, or x² + 2x + 2 = 0. It is a quadratic equation even though it looks cubic.
Q5. The product of two consecutive positive integers is 306. Write this as a quadratic equation.
Let the first integer be x, so the next is x + 1. x(x + 1) = 306, that is x² + x − 306 = 0.
Q6. The area of a rectangular plot is 528 m² and its length is one more than twice its breadth. Write this as a quadratic equation.
Let the breadth be x and the length 2x + 1. Then x(2x + 1) = 528, that is 2x² + x − 528 = 0.
Q7. What is a root of a quadratic equation? How many roots can there be?
A real number α for which aα² + bα + c = 0. The zeroes of the quadratic polynomial and the roots of the equation are the same, and there are at most two roots.
Q8. Solve by factorisation: x² − 3x − 10 = 0.
Splitting the middle term: x² − 5x + 2x − 10 = (x − 5)(x + 2) = 0. So x = 5 or x = −2.
Q9. Solve by factorisation: 2x² + x − 6 = 0.
2x² + 4x − 3x − 6 = (2x − 3)(x + 2) = 0. So x = 3/2 or x = −2.
Q10. Solve by factorisation: √2 x² + 7x + 5√2 = 0.
(√2 x + 5)(x + √2) = 0. So x = −5/√2 or x = −√2.
Q11. Solve by factorisation: 100x² − 20x + 1 = 0.
(10x − 1)² = 0, so x = 1/10 (both roots are equal).
Q12. A prayer hall has an area of 300 m² and its length is one more than twice its breadth. Find its dimensions.
Let the breadth be x, then 2x² + x − 300 = 0, which factorises as (x − 12)(2x + 25) = 0. x = 12 or −12.5, but a breadth cannot be negative. So the breadth is 12 m and the length is 2 × 12 + 1 = 25 m.
Q13. Find two numbers whose sum is 27 and product is 182.
Let one number be x, then the other is 27 − x. x(27 − x) = 182, that is x² − 27x + 182 = 0, whose roots are 13 and 14. The numbers are 13 and 14.
Q14. Find two consecutive positive integers the sum of whose squares is 365.
Take x and x + 1: x² + (x + 1)² = 365, that is x² + x − 182 = 0, whose roots are 13 and −14. Being positive, x = 13. The integers are 13 and 14.
Q15. The 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 be x, then the altitude is x − 7. x² + (x − 7)² = 13², that is x² − 7x − 60 = 0, whose roots are 12 and −5. Taking the positive value x = 12. The base is 12 cm and the altitude is 5 cm.
Q16. What is the discriminant and how does it tell the nature of the roots? Also write the quadratic formula.
The discriminant is D = b² − 4ac. D > 0: two distinct real roots. D = 0: two equal real roots. D < 0: no real roots. The roots are x = (−b ± √(b² − 4ac)) / 2a (when b² − 4ac ≥ 0).
Q17. Find the discriminant of 2x² − 4x + 3 = 0 and state the nature of its roots.
a = 2, b = −4, c = 3. D = (−4)² − 4 × 2 × 3 = 16 − 24 = −8 < 0. So there are no real roots.
Q18. State the nature of the roots of 2x² − 6x + 3 = 0 and find them.
D = 36 − 24 = 12 > 0, so there are two distinct real roots. x = (6 ± √12)/4 = (3 ± √3)/2.
Q19. Find the values of k for which 2x² + kx + 3 = 0 has two equal roots.
For equal roots D = 0: k² − 4 × 2 × 3 = 0, that is k² = 24, so k = ±2√6.
Q20. Is it possible to design a rectangular park of perimeter 80 m and area 400 m²?
Length + breadth = 40. If the length is x, then x(40 − x) = 400, that is x² − 40x + 400 = 0. D = 1600 − 1600 = 0, so the roots are real and equal and it is possible. x = 20, so length = breadth = 20 m (a square park).
Quiz
1. What are the roots of x² − 3x − 10 = 0?
(x − 5)(x + 2) = 0, so x = 5 or −2.
2. What is the discriminant of 2x² − 4x + 3 = 0?
D = 16 − 24 = −8.
3. If b² − 4ac = 0, what are the roots of the quadratic equation like?
When D = 0 there are two equal real roots.
4. What is the maximum number of roots a quadratic equation can have?
An equation of degree 2 has at most two roots.
5. ax² + bx + c = 0 is a quadratic equation only when...
If a = 0 the x² term disappears, so a ≠ 0 is needed.