Exercise 10.4.3

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

Note:we can use the discriminant Δ of the quadratic equation ax2+bx+c=0. The discriminant is given by Δ=b^2−4ac.

If Δ>0, the quadratic equation has two distinct real roots.

If Δ=0, the quadratic equation has one real root (a repeated root).

If Δ<0, the quadratic equation has no real roots (the roots are complex).

(i) 2x2 – 3x + 5 = 0

The coefficients are a=2,b=-3,c=5

Calculate the discriminant = b2-4ac

= −32 - 4.2.5

= 9 - 40 = .

(ii) 3x2-43x+4=0

The coefficients are a = 3,b= -43, c=4

Calculate the discriminant = b2-4ac

= −432 - 4.3.4

= -/.

(iii) 2x2-6x+3=0

Solutions:

The coefficients are a = 2,b= -6, c=3

Calculate the discriminant = b2-4ac

= −62 - 4.2.3

= .

Since Δ>0, the equation 2x2-6x+3=0 has two distinct real roots.

To find the roots, use the quadratic formula: x = .

So, the roots of the equation 2x2-6x+3=0 are

x= .

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

(i) 2x2 + kx + 3 = 0

Solution:

The coefficients are a=2,b=k,c=3

Calculate the discriminant Δ:Δ=b^2−4ac

Δ = k2 - 24

For the equation to have equal roots, Δ must equal zero: k2 -24=0

Therefore, the values of k for which the equation 2x2 + kx + 3 = 0 has two equal roots are k = 2√6 and k-=-2√6.

(ii) kx (x – 2) + 6 = 0

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