Revision: Quadratic Equations Algebra Maths 1 SSC (English Medium) 10th Standard Maharashtra State Board

An equation with one variable, in which the highest power of the variable is two, is known as a quadratic equation.

Standard Form:

 ax² + bx + c = 0,  a ≠ 0

For example :

(i) 3x² + 4x + 7 = 0

(ii) 4x² + 5x = 0 

The set of elements representing the roots of a quadratic equation is called its solution set.

A value of the variable which satisfies the equation is called a root (solution).

If substituting a value of x makes the polynomial zero, that value is a root.

• A number α is called a root of ax² + bx + c = 0, if aα² + bα + c = 0

If a quadratic equation contains only two terms where one is a square term and the other is the first power term of the unknown, it is called adjected quadratic equation.

For example :

(i) 4x² + 5x = 0

(ii) 7x² − 3x = 0, etc. 

If the quadratic equation contains only the square of the unknown, it is called a pure quadratic equation.

For example :

(i) x² = 4 

(ii) 3x² − 8 = 0, etc.

For the quadratic equation ax² + bx + c = 0, a ≠ 0; the expression b² − 4ac is called the discriminant and is, in general, denoted by the letter 'D'.

Thus, discriminant D = b² − 4ac.

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

For a quadratic polynomial

ax² + bx + c (a≠0)

If its zeroes are α and β, then:

\[\alpha+\beta=-\frac{b}{a}\]

\[\alpha\beta=\frac{c}{a}\]

For a cubic polynomial

ax³ + bx² + cx + d, 

\[\alpha+\beta+\gamma=-\frac{b}{a}\]

\[\alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a}\]

\[\alpha\beta\gamma=-\frac{d}{a}\]

The quadratic equation whose roots are α and β is

x² − (α+β)x + αβ = 0

Given the roots of the equation (q – r)x² + (r – p)x + (p – q) = 0 are equal.

∴ Discriminant (D) = 0

⇒ b² – 4ac = 0

⇒ (r – p)² – 4 × (q – r) × (p – q) = 0

⇒ r² + p² – 2pr – 4[qp – q² – rp + qr] = 0

⇒ r² + p² – 2pr – 4qp + 4q² + 4rp – 4qr = 0

⇒ r² + p² + 2pr – 4qp – 4qr + 4q² = 0

⇒ (p + r)² – 4q(p + r) + 4q² = 0

Let (p + r) = y

⇒ y² – 4qy + 4q² = 0

⇒ (y – 2q)² = 0

⇒ y – 2q = 0

⇒ y = 2q

⇒ p + r = 2q

Hence proved.

In the factorisation method, the quadratic expression is written as a product of two linear factors

1. Clear fractions and brackets, if any.

2. Transpose all terms to one side to get the standard form
   ax² + bx + c = 0
3. Factorise the quadratic expression into two linear factors.

4. Put each factor equal to zero (using the zero product rule).

5. Solve the resulting linear equations to obtain the roots.

The completing the square method is used when a quadratic equation cannot be factorised easily.

1. First, write the equation in standard form.
   $ax2+bx+c=0$

2. If a ≠ 1, divide the entire equation by a.

3. Add and subtract \[\left(\frac{b}{2}\right)^2\] to complete the square.

4. Convert the left-hand side into a perfect square.

5. Solve the resulting equation to find the roots.

1. Write the given equation in the standard form

   ax² + bx + c = 0

2. Identify the values of a, b, and c.

3. Find the value of the discriminant

   D = b² − 4ac

4. Substitute the values of a, b, and D in the formula

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

5. Simplify to obtain the roots.

D = b² – 4ac