Nature of Roots of Quadratic Equation NCERT Exercise 4.4 part one Class Ten Mathematics

## Exercise 4.4 Part 1

Question 1: Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

Question (i) 2x^2 – 3x + 5

Solution: We have a = 2, b = - 3 and c = 5

D=b^2-4ac

=(-3)^2-4xx2xx5

=9-40=-31

Here; D < 0; hence no real root is possible.

Question (ii): 3x^2 - 4sqrt3 x + 4 = 0

Solution: We have; a = 3, b = - 4sqrt3 and c = 4

D=b^2-4ac

=(-4sqrt3)^2-4xx3xx4

=48-48=0

Here; D = 0; hence root are equal and real.

Root can be calculated as follows:

text(Root)=(-b)/(2a)

=(4sqrt3)/(6)=(2sqrt3)/(3)

Question (iii) 2x^2 – 6x + 3 = 0

Solution: We have; a = 2, b = - 3 and c = 3

D=b^2-4ac

=(-6)^2-4xx2xx3

=36-24

Here; D > 0; hence roots are real and different

Now, roots can be calculated as follows:

α=(-b+sqrtD)/(2a)

=(6+sqrt(12))/(2xx2)=(6+2sqrt3)/(4)

=(3+sqrt3)/(4)

β=(-b-sqrt3)/(2a)

=(6-2sqrt3)/(4)=(3-sqrt3)/(2)

Question 2: Find the value of k for each of the following quadratic equations, so that they have two equal roots.

Question (i) 2x^2 + kx + 3 = 0

Solution: We have; a = 2, b = k and c = 3

For equal roots; D should be zero.

Hence;

b^2-4ac=0

Or, k^2-4xx2xx3=0

Or, k^2-24=0

Or, k^2=24

Or, k=2sqrt6

Question (ii) kx(x – 2) + 6 = 0

Solution: kx(x – 2) + 6 = 0

Or, kx^2 – 2kx + 6 = 0

Here; a = k, b = - 2k and c = 6

For equal roots, D should be zero

b^2-4ac=0

Or, (-2k)^2-4xxkxx6=0

Or, 4k^2-24k=0

Or, k^2=6k

Or, k=6