# Nature and sign of the roots of a quadratic equation

**Nature of the Roots**

As I mentioned earlier in the post “Roots of a quadratic equation” that Discriminant helps in determining the nature of the roots of a quadratic equation ax^2 + bx + c =0.

The expression (b^2 – 4ac) under the radical sign is called the **Discriminant** of equation.

**1.)** If discriminant is **positive ** (that is, if b^2 – 4ac > 0), then the roots p and qof equation are **real ** and **unequal.**

*1.a.)* If discriminant is **positive ** and a **perfect square **then the roots of equation are **real, rational and unequal**

*1.b.)* If the discriminant is **positive **but **not a perfect square** then the roots of equation are **real, irrational and unequal.**

**2.)** If discriminant is **zero** (that is, if b^2 – 4ac = 0), then the roots p and q of equation are **real ** and **equal. **

**3.)** If discriminant is **negative **(that is, if b^2 – 4ac < 0), then the roots p and ß of equation are **imaginary **and **unequal.**

**4.)** If b^2 – 4ac is a **perfect square **but any one of **a **or **b **is irrational then the roots of equation are irrational.

If the roots of a quadratic equation are irrational (a,b,c being rational) they will of the form u+sqrt v and u-sqrt v i.e. they will occur as a conjugate pair.

Q. Discuss the nature of the roots of the equation x^2 – 26x + 169 = 0 ?

The discriminant of the quadratic equation x^2 – 26x + 169 = 0 is D =(-26)^{2} – 4.1.169 = 676 – 676 = 0

So the discriminant of the given eq. is 0 and the coefficients of x^{2}, x are rational, hence roots of the equation are real, rational and equal.

Q. If a, b, c are rational and a + b + c = 0, show that the roots of the equation ax^{2} + bx + c = 0 are rational.

a + b + c = 0 –> b = -(c + a)

The discriminant of equation (1) is b^2-4ac= [-(c + a)]^2 – 4ac = (c + a)^2 – 4ac = (c – a)^2

Since a, b, c are rational and the discriminant of equation is a perfect square,

hence the roots of (1) are rational.

**Sign of the roots**

On the basis of sum of roots and product of roots of the quadratic equation we can determine whether the roots are +ve or -ve. Following table below shows how to determine the sign of the roots of a quadratic equation.

**sign of product of roots sign of sum of roots sign of roots **

+ve +ve Both roots are +ve

+ve -ve Both roots are -ve

-ve +ve Numerically larger root is +ve and 2nd root is -ve

-ve -ve Numerically larger root is -ve and 2nd root is +ve

Q. Discuss the sign of the roots of the equation x^2 – 26x + 169 = 0 ?

sum of roots =-( -26/1) = 26 = +ve, product of roots = 169/1 =169 = +ve

so, Both roots are +ve.

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