Question

Question 2 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

Class 10 - Math - Quadratic Equations Page 91

Answer 1

(i) 2x2 + kx + 3 = 0
Comparing equation with ax2 + bx + c = 0, we get
a = 2, b = k and c = 3
Discriminant = b2 - 4ac

= (k)2 - 4(2) (3)
= k2 - 24
For equal roots,
Discriminant = 0
k2 - 24 = 0
k2 = 24
k = ±√24 = ±2√6

(ii) kx(x - 2) + 6 = 0
or kx2 - 2kx + 6 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = k, b = - 2k and c = 6
Discriminant = b2 - 4ac
= ( - 2k)2 - 4 (k) (6)
= 4k2 - 24k
For equal roots,
b2 - 4ac = 0
4k2 - 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 'x2' and 'x'.
Therefore, if this equation has two equal roots,k should be 6 only

Answer 2

I) 2x² + kx + 3 = 0

here,

a = 2 , b = k , c = 3

°.° mentioned equations have two equal roots.

.°. b² - 4ac = 0

=> k² - 4 × 2 × 3 = 0

=> k² - 24 = 0

=> k = ±√24

=> k = ± 2√6

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ii) kx( x - 2 ) + 6 = 0

here,

a = k , b= -2k, c = 6

°.° mentioned equations have two equal roots

. °. b² - 4ac

=> ( - 2k )² - 4 × k × 6 = 0

=> 4k² - 24k = 0

=> 4k ( k - 6 ) = 0

Either

4k = 0

=> k = 0

or

k - 6 = 0

=> k = 6

.°. k = 0,6

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