Question

Find the values of k for each of the following quadratic equations, so that they have two equal roots. 
(i) \(2x^2 + kx + 3 = 0\) (ii) \(kx (x – 2) + 6 = 0\)

Answer 1

We know that if an equation \(ax^2 + bx + c = 0\) has two equal roots, its discriminant \((b^2 − 4ac)\) will be 0.

(i)  \(2x^2 + kx + 3 = 0\)

Comparing equation with \(ax^2 + bx + c = 0,\) we obtain
a = 2, b = k, c = 3

Discriminant = \(b^2 − 4ac\) = \((k)^2− 4(2) (3)\) =\(k^2 − 24\)
For equal roots, Discriminant = 0 
\(k^2 − 24\) = 0 
\(k^2 = 24\)
\(k = ±\sqrt{24} = ±2\sqrt6\)

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(ii) \(kx (x − 2) + 6 = 0\)
or \(kx^2 − 2kx + 6 = 0\)

Comparing this equation with \(ax^2 + bx + c = 0,\) we obtain
a = k, b = −2k, c = 6

Discriminant = \(b^2 − 4ac\) = \((− 2k)^2 − 4 (k)\)\((6)\) = \(4k^2 − 24k\)
For equal roots, 
\(b^2 − 4ac\)= 0 
\(4k^2 − 24k\) = 0 
\(4k (k − 6)\)= 0 
Either 4k = 0 or k = 6 = 0 
k = 0 or k = 6 

However, if k = 0, then the equation will not have the terms \(‘x^2’\) and ‘x’.

Therefore, if this equation has two equal roots, k should be 6 only.