LEARN MATH STEP BY STEP THROUGH VERY EASY PROCESS

NATURE OF A ROOTS OF A QUADRATIC EQUATIONS

**Nature of Roots of a Quadratic Equations –
**

**Consider the quadratic equations ax²+ bx + c = 0, where a, b, c are real numbers and a ≠ 0. Let α and β be the roots of this equation.
**

**Let, D = (b²- 4ac). Then, D is called the discriminant of the given equation.
**

**A. If D > 0, the roots area real and distinct, given by
**

** - b + √D - b - √D
**

** α = ------------ and β = -------------
**

** 2a 2a
**

__Case 1.__ If D > 0 and not a perfect square, then the roots are unequal and irrational.

__Case 2.__ If D > 0 and a perfect square, then the roots are –

**(i) Rational and unequal, if a, b, c are rational
**

**(ii) Irrational and unequal, if b is irrational**

** - b**

**B. If D = 0, the roots are real and equal, each equal to ------.
**

** 2a
**

**C. If D < 0, the roots are unequal and imaginary.
**

** Note : Clearly, for real roots, we must have, D ≥ 0.**

**Example.1) without solving following equations, find the value of ‘p’ for which the roots are real and equal. px²- 4x + 3 = 0
**

**Ans.) The given equation is px²- 4x + 3 = 0
**

**This is of the form ax²+ bx + c = 0, where a = p, b = - 4, and c = 3.
**

**So, D = b²- 4ac = (- 4)²- 4 X p X 3
**

** = 16 – 12p
**

**For the given equation to have real and equal roots, we must have D = 0.
**

**Now, D = 0
**

**=> 16 – 12p = 0
**

**=> 12p = 16
**

**=> p = 16/12 = 4/3
**

**Hence, the required value of p is 4/3. (Ans.)**

**Example.2) Without solving the given quadratic equation, find the value of ‘p’ for which it has real and equal roots => x²+ (p – 3)x + p = 0
**

**Ans.) The given equation is x²+ (p – 3)x + p = 0
**

**This is of the form ax²+ bx + c = 0, where a = 1, b = (p – 3), c = p
**

**So, D = (b²- 4ac)
**

**For real & equal roots, we must have D = 0
**

**So, D = 0 => {(p – 3)²- (4 X 1 X p)} = 0
**

**=> (p²- 6p + 9 – 4p) = 0
**

**=> p²- 10p + 9 = 0
**

**=> p²- (9 + 1)p + 9 = 0
**

**=> p²- 9p – p + 9 = 0
**

**=> p(p – 9) – (p – 9) = 0
**

**=> (p – 9) (p – 1) = 0
**

**=> (p – 9) = 0 or (p – 1) = 0
**

**=> p = 9 or p = 1
**

**Hence, p = 9 or p = 1 (Ans.)**

**Example.3) Without solving the following quadratic equation, find the set of value of m for which the given equation has real and equal roots, x²+ 2(m -1)x + (m + 5) = 0
**

**Ans.) The given equation is x²+ 2(m -1)x + (m + 5) = 0
**

**This is of the form of ax²+ bx + c = 0, where a = 1, b = 2(m – 1), and c = (m + 5)
**

**As we know D = (b²- 4ac)
**

**So, D = [{2(m – 1)}² - {4 X 1 X (m + 5)}]
**

** = {4(m²- 2m +1) – (4m + 20)}
**

** = (4m²- 8m + 4 – 4m – 20) = 4m²- 12m – 16
**

**For real & equal roots, we must have D = 0
**

**Now, D = 0, so
**

**=> 4m²- 12m – 16 = 0
**

**=> 4m²- (16 – 4)m – 16 = 0
**

**=> 4m²- 16m + 4m – 16 = 0
**

**=> 4m(m – 4) + 4(m – 4) = 0
**

**=> (m – 4) (4m + 4) = 0
**

**=> 4(m – 4)(m + 1) = 0
**

**=> (m – 4)(m + 1) = 0
**

**=> (m – 4) = 0 or (m + 1) = 0
**

**=> m = 4, or m = - 1
**

**Hence, the required values of m is 4 or -1 and the required set is {4, -1} (Ans.)**

**Example.4) Show that the equation 3x²+ 7x + 8 = 0 is not true for any real value of x
**

**Ans.) The given equation is 3x²+ 7x + 8 = 0. This is of the form ax²+ bx + c = 0, where a = 3, b = 7, c = 8
**

**As we know, D = (b²- 4ac) = {(7)²- (4 X 3 X 8)}
**

** = (49 – 96) = - 47 < 0
**

**So, the given equation has no real roots.
**

**So, 3x²+ 7x + 8 = 0 is not true for any real value of x (Ans.)**

**Example.5) Without solving the given equation comment upon the nature of its roots
**

**(i) 3x²- 11x + 10 = 0, **

**Ans.) The given equation is 3x²- 11x + 10 = 0
**

**This is the form of ax²+ bx + c = 0, where a = 3, b = -11, c = 10
**

**As we know, D = (b²- 4ac) = {(- 11)² - (4 X 3 X 10)}
**

** = (121 – 120) = 1 > 0
**

**Thus, D > 0 and D is not a perfect square.
**

**Hence, the roots are unequal and irrational. (Ans.)**

**(ii) 15 x²- 12x + 3 = 0,**

**The given equation is 15 x²- 12x + 3 = 0
**

**This is the form of ax²+ bx + c = 0, where a = 15, b = -12, c = 3
**

**As we know, D = (b²- 4ac) = {(-12)²- (4 X 15 X 3)}
**

** = (144 – 180) = - 36 < 0
**

**Thus, D < 0, hence the given equation has no real roots. (Ans.)**

**(iii) 9x²- 6x + 1 = 0**

**The given equation is 9 x²- 6x + 1 = 0
**

**This is the form of ax²+ bx + c = 0, where a = 9, b = -6, c = 1
**

**As we know, D = (b²- 4ac) = {(-6)² - (4 X 9 X 1)}
**

** = (36 – 36) = 0
**

**Hence, the roots of the given equation are real and equal. (Ans.)**

Your second block of text...