Students can go through AP SSC 10th Class Maths Notes Chapter 5 Quadratic Equations to understand and remember the concepts easily.

## AP State Syllabus SSC 10th Class Maths Notes Chapter 5 Quadratic Equations

→ The general form of a linear equation in one variable is ax + b = c.

→ Any equation of the form p(x) = 0 where p(x) is a polynomial of degree 2, is a quadratic equation.

→ If p(x) = 0 whose degree is 2 is written in descending order of their degrees, then we say that the quadratic equation is written in the standard form.

→ The standard form of a quadratic equation is ax^{2} + bx + c = 0 where a ≠ 0. We can write it as y = ax^{2} + bx + c.

→ There are various occasions in which we make use of Q.E. in our day-to-day life.

Eg : The height of a rocket is defined by a Q.E.

→ Let ax^{2} + bx + c = 0 be a quadratic equation. A real number α is called a root of the Q.E. if aα^{2} + bα + c = 0. And x = α is called a solution of the Q.E.

(i.e.) the real value of x for which the Q.E ax^{2} + bx + c = 0 is satisfied is called its solution.

→ Zeroes of the Q.E. ax^{2} + bx + c = 0 and the roots of the Q.E. ax^{2} + bx + c = 0 are the same.

→ To factorise a Q.E. ax^{2} + bx + c = 0, we find p, q ∈ R such that p + q = b and pq = ac. This process is called Factorising a Q.E. by splitting its middle term.

Eg : 12x^{2} + 13x + 3 = 0

here a = 12; b = 13; c = 3

a.c = 12 × 3 = 36 = 4 × 9 and

b = 4 + 9 here p = 9 and q = 4

Now 12x^{2} + 13x + 3 = 0

⇒ 12x^{2} + 9x + 4x + 3 = 0

⇒ 3x(4x + 3) + 1 (4x + 3) = 0

⇒ (4x + 3) (3x + 1) = 0

Here 4x + 3 = 0 or 3x + 1 = 0

⇒ 4x = -3 or 3x = -1

⇒ x = \(\frac{-3}{4}\) or \(\frac{-1}{3}\)

\(\frac{-3}{4}\) and \(\frac{-1}{3}\) are called the roots of the Q.E. 12x^{2} + 13x + 3 = 0 and x = \(\frac{-3}{4}\) or \(\frac{-1}{3}\) is the solution of the Q.E. 12x^{2} + 13x + 3 = 0.

→ In the above example, (4x + 3) and (3x +1) are called the linear factors of the Q.E. 12x^{2} + 13x + 3 = 0.

→ We can factorise a Q.E. by adjusting its left side so that it becomes a perfect square.

Eg: x^{2} + 6x + 8 = 0

⇒ x^{2} + 2.x.3 + 8 = 0

⇒ x^{2} + 2.x.3 = -8

The L.H.S. is of the form a^{2} + 2ab

∴ By adding b^{2} it becomes a perfect square

∴ x^{2} + 2.x.3 + 3^{2} = -8 + 3^{2}

⇒ (x + 3)^{2} = -8 + 9

⇒ (x + 3)^{2} = 1

⇒ x + 3 = ± 1 Now we take x + 3 = 1 or x + 3 = -l

⇒ x = -2 or x = -4

→ Adjusting a Q.E. of the form ax^{2} + bx + c = 0 so that it becomes a perfect square.

Step – 1 : ax^{2} + bx + c = 0 ⇒ ax^{2} + bx = -c ⇒ x^{2} + \(\frac{b}{a}\)x = \(\frac{-c}{a}\)

Step – 2 :

Step – 3 :

Step – 4 : Solve the above.

E.g: 5x^{2} – 6x + 2 = 0

→ Let ax^{2} + bx + c = 0 (a ≠ 0) be a Q.E., then b^{2} – 4ac is called the Discriminant of the Q.E.

→ If b^{2} – 4ac > 0, then the roots of the Q.E. ax^{2} + bx + c = 0 are given by

x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\). This is called quadratic formula to find the roots.

The nature of the roots of a Q.E. can be determined either by its discriminant or its graph.

Q.E.: y = ax^{2} + bx + c.