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 x2, 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 ax2 + 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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