ICSE Class 10 Maths Chapter 06 Quadratic Equations

Chapter 6

Quadratic Equations

Points To Remember

1. Quadratic Equation: An equation of the form \(ax^2 + bx + c = 0\), where a, b, c are real numbers and a ≠ 0, is called a quadratic equation.

2. Roots of Quadratic Equation: A number say α is called a root of the equation \(ax^2 + bx + c = 0\) if \(aα^2 + bα + c = 0\). Value of α will satisfy the given quadratic equation.

3. Zero Product Rule: Let a and b be the two real numbers, then ab = 0 ⇒ a = 0 or b = 0 or both equal to zero. This is called zero product rule.

4. Method for solving the quadratic equation: (A) By Factorization:

(i) Make the given equation free from fractions and radicals if any and then put it into the standard form \(ax^2 + bx + c = 0\).

(ii) Now factorise \(ax^2 + bx + c\) into two linear factors.

(iii) Put each factor equal to zero (by zero product rule)

(iv) Now solve these linear equations, then we shall get two roots of the given quadratic equation.

(B) By Quadratic formula:

Let α and β be the roots of the quadratic equation \(ax^2 + bx + c = 0\), where a ≠ 0, then

\[α = \frac{-b + \sqrt{b^2 - 4ac}}{2a}\]

and

\[β = \frac{-b - \sqrt{b^2 - 4ac}}{2a}\]

Proof. \(ax^2 + bx + c = 0\)

⇒ \(4a^2x^2 + 4abx + 4ac = 0\) (Multiplying by 4a)

⇒ \(4a^2x^2 + 4abx = -4ac\)

Adding \(b^2\) both sides, we get

\(4a^2x^2 + 4abx + b^2 = b^2 - 4ac\)

⇒ \((2ax)^2 + 2 × 2ax·b + (b)^2 = b^2 - 4ac\)

⇒ \((2ax + b)^2 = b^2 - 4ac\)

⇒ \(2ax + b = ± \sqrt{b^2 - 4ac}\) (Taking square root on both sides)

\[x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}\]

If α and β be the roots of the given quadratic equation, then

\[α = \frac{-b + \sqrt{b^2 - 4ac}}{2a}\]

and

\[β = \frac{-b - \sqrt{b^2 - 4ac}}{2a}\]

Note. \(b^2 - 4ac\) is called the determinant of the quadratic equation \(ax^2 + bx + c = 0\) and is denoted by D.

∴ D = \(b^2 - 4ac\)

5. Some different forms of the quadratic equation

(i) Equations of the form:

\[\sqrt{ax + b} = (cx + d)\]

We can find their solutions if \(ax + b ≥ 0\) and \(cx + d ≥ 0\).

Note. (i) We shall solve it first by squaring and then writing in the standard form.

(ii) We shall take only positive value of the square root.

(ii) Equations of the form

\[\sqrt{ax^2 + bx + c} = (dx + c)\]

We can find their solutions if \(ax^2 + bx + c ≥ 0\)

Note. (i) We shall solve it first by squaring it and then write it in the standard form.

(ii) We shall take only positive value of square root.

Exercise 6 (A)

Q. 1. Find which of the following are the solutions of the equation \(6x^2 - x - 2 = 0\)?

(i) \(\frac{1}{2}\) (ii) \(-\frac{1}{2}\) (iii) \(\frac{2}{3}\)

Sol. Given equation is \(6x^2 - x - 2 = 0\)

(i) If \(x = \frac{1}{2}\) is its solution, then it will satisfy the equation. Now, substituting the value of \(x = \frac{1}{2}\) is the given equation.

Substituting \(x = \frac{1}{2}\)

\[= 6 \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) - 2\]

\[= 6 × \frac{1}{4} - \frac{1}{2} - 2\]

\[= \frac{3}{2} - \frac{1}{2} - 2 = \frac{3 - 1 - 4}{2} = \frac{-2}{2} = -1 ≠ 0\]

∴ \(x = \frac{1}{2}\) is not a root of the equation

\(6x^2 - x - 2 = 0\)

(ii) Given equation is \(6x^2 - x - 2 = 0\)

Substituting \(x = -\frac{1}{2}\) in \(6x^2 - x - 2 = 0\)

\[6\left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right) - 2 = 0\]

\[6 × \frac{1}{4} + \frac{1}{2} - 2 = 0\]

\[\frac{3}{2} + \frac{1}{2} - 2 = 0\]

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