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`Copyright © excellup 2014