CBSE Class 10 Mathematics Quadratic Equations Notes Set 11

Quadratic Equation and Its Roots

A quadratic equation in x has the general form ax² + bx + c = 0 with real a, b, c and a ≠ 0. If b = 0 it is of the type ax² + c = 0, if c = 0 it is ax² + bx = 0, and if both are zero it is ax² = 0.
A value of x that satisfies the equation is a root (solution). A quadratic equation has at most two roots. A linear equation results if a = 0, b ≠ 0; if a = b = c = 0 every x satisfies it.

Methods of Solving

1. Factorisation: split the middle term so that the product of the two parts equals ac, factorise and equate each factor to zero.
2. Completing the square: divide by a, move the constant to the right side, add (b/2a)² to both sides and take square roots.
3. Quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which comes from completing the square on the general equation.

Discriminant and Nature of Roots

D = b² − 4ac is the discriminant.
1. D > 0: two real and distinct roots. If a, b, c are rational and D is a perfect square the roots are rational; if D is not a perfect square they are irrational.
2. D = 0: two real and equal roots, each equal to −b/2a.
3. D < 0: no real roots (the roots are complex conjugates).

Sum and Product of Roots

If α and β are the roots, then α + β = −b/a and αβ = c/a. The equation with roots α and β is x² − (α + β)x + αβ = 0.
If a, b, c have the same sign, both roots (when real) are negative. If a and c have the same sign and b has the opposite sign, both roots (when real) are positive.

Word Problems

Convert the story into an equation in one unknown, solve, and discard roots that are impossible (negative lengths, negative counts and so on).
Useful forms: two consecutive integers x, x + 1; two consecutive odd or even integers x, x + 2; numbers with a fixed sum S: x and S − x; time = distance ÷ speed; number of items = total cost ÷ price per item.

Solved Examples

Example 1. Find the roots of x² − x − 3 = 0.
Answer:
a = 1, b = −1, c = −3, so D = 1 + 12 = 13 > 0.
\( x = \frac{1 \pm \sqrt{13}}{2} \). The roots are irrational and distinct.

Example 2. For what value of K is x = −a a solution of x² − 2(a + b)x + 3K = 0?
Answer:
Substitute: a² + 2a(a + b) + 3K = 0, so 3K = −3a² − 2ab.
K = −a(3a + 2b)/3

Example 3. Solve x² − 2ax + a² − b² = 0 by factorisation.
Answer:
The expression is (x − a)² − b² = (x − a − b)(x − a + b).
x = a + b or x = a − b

Example 4. Solve x² + 8x + 7 = 0.
Answer:
Split the middle term: x² + 7x + x + 7 = (x + 7)(x + 1) = 0.
x = −7 or x = −1

Example 5. Solve 16x² − 24x = 0.
Answer:
8x(2x − 3) = 0.
x = 0 or x = 3/2

Example 6. Solve \( \frac{2x}{x-3} + \frac{1}{2x+3} + \frac{3x+9}{(x-3)(2x+3)} = 0 \).
Answer:
Here x ≠ 3 and x ≠ −3/2. Multiply by (x − 3)(2x + 3): 2x(2x + 3) + (x − 3) + 3x + 9 = 0.
4x² + 10x + 6 = 0, so 2x² + 5x + 3 = 0 and (2x + 3)(x + 1) = 0.
x = −3/2 is not allowed, so x = −1.

Example 7. Solve 25x² − 30x + 9 = 0.
Answer:
(5x − 3)² = 0.
x = 3/5 (repeated root)

Example 8. Solve x² + 3x + 1 = 0 by completing the square.
Answer:
x² + 3x + 9/4 = 9/4 − 1 = 5/4, so (x + 3/2)² = 5/4.
x = (−3 ± √5)/2

Example 9. Solve 3x² − 2√15 x − 2 = 0 by completing the square.
Answer:
Divide by 3: x² − (2√15/3)x = 2/3. Add (√15/3)² = 15/9: (x − √15/3)² = 21/9.
x = (√15 ± √21)/3

Example 10. Solve x² − 7x − 5 = 0.
Answer:
D = 49 + 20 = 69 > 0.
x = (7 ± √69)/2

Example 11. If the roots of a(b − c)x² + b(c − a)x + c(a − b) = 0 are equal, show that 2/b = 1/a + 1/c.
Answer:
Equal roots: b²(c − a)² − 4a(b − c)c(a − b) = 0, which simplifies to (ab + bc − 2ac)² = 0.
So ab + bc = 2ac; dividing by abc gives 1/c + 1/a = 2/b. Proved.

Example 12. If the roots of x² − 8x + a² − 6a = 0 are real and distinct, find the possible values of a.
Answer:
D > 0: 64 − 4(a² − 6a) > 0, so a² − 6a − 16 < 0, i.e. (a − 8)(a + 2) < 0.
−2 < a < 8

Example 13. The sum of the squares of two consecutive positive integers is 545. Find them.
Answer:
x² + (x + 1)² = 545 gives x² + x − 272 = 0, i.e. (x − 16)(x + 17) = 0.
The integers are 16 and 17.

Example 14. The sum of two numbers is 48 and their product is 432. Find the numbers.
Answer:
x(48 − x) = 432 gives x² − 48x + 432 = 0, i.e. (x − 36)(x − 12) = 0.
The numbers are 12 and 36.

Example 15. The length of a hall is 5 m more than its breadth and its area is 84 m². Find its dimensions.
Answer:
x(x + 5) = 84 gives x² + 5x − 84 = 0, so (x + 12)(x − 7) = 0 and x = 7 (positive).
Breadth 7 m and length 12 m.

Example 16. Out of a group of swans, 7/2 times the square root of the total number are playing on the shore and the remaining 2 are in deep water. Find the total number of swans.
Answer:
x = (7/2)√x + 2, so x − 2 = (7/2)√x. Squaring: 4x² − 65x + 16 = 0, i.e. (x − 16)(4x − 1) = 0.
x = 1/4 is rejected (and fails the original equation). 16 swans.

Example 17. Find n if the sum of the first n natural numbers is 276.
Answer:
n(n + 1)/2 = 276 gives n² + n − 552 = 0, so (n − 23)(n + 24) = 0.
n = 23