1.

Find the value of k for which the Givenequation has real and equal oroots .1)2x² - 10x+k=0 2) kx²5x+K=03) x^2+ k(4x+k-1) +2=04) x^2-2k( 1+3k] +7 (3+2k),,,5)kx(x-3)+9=0 ..6)kx(x- 2√5)+10=0​

Answer»

Step-by-step EXPLANATION:

solution:>

for equal and REAL ROOTS => b²= 4ac

(1) (-10)²= = 8k => k= 100/8= 12.5

(2) (±5)²= 4k² => k²= 25/4=> k= ±5/2

(3) (4k)²= k²-k+2 => 15k²+k-2=0

15k²+6k-5k-2=0

3k(5 k+2) - 1( 5k+2)=0

(5k+2) ( 3k-1)= 0=> k = - 2/5 and k= 1/3

(4) here b=0

4ac= 4[ -2k(1+3k)+7(3+2k)

=> 0 = 4[ -2k-6k²+21+14k]

=>. 4[ -6k²+12k+21]= 0

=> 2k²-4k-7= 0 dividing by - 12

k= [ 4±√(16-4×2×-7)] / 4

= ( 4± √-40)/4=> k= 1± (√-10)/2

(5) b²= (-3k)²= 9k²

4ac= 4×k×9= 36k

=> 9k²= 36k => 9k( k-4)=0

=> k=0 or k=4

(6) b²= (-2√5)²= 20,. 4ac= 4×k×10= 40K

=>. 20= 40k=>. k= 1/2


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