For what values of k will the equation{{x}^{2}}-2(1+3k)x+
7(3+2k)=0
have equal roots [MP PET 1997]
A) 1,-\frac{10}{9} B) 2,-\frac{10}{9} C) 3,-\frac{10}{9} D) 4,-\frac{10}{9}
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Hi ,
compare x² - 2( 1 + 3k )x + 7( 3 + 2k ) = 0
with ax² + bx + c = 0 we get ,
a = 1 ,
b = - 2 ( 1 + 3k ) ,
c = 7 ( 3 + 2k ) ,
discreaminant = 0
[ since roots are equal ]
b² - 4ac = 0
[ - 2 ( 1 + 3k ) ]² - 4 × 1 × [ 2 ( 1 + 3k ) ] = 0
[2 ( 1 + 3k ]² - 4 [ 2 ( 1 + 3k ) ] = 0
4 ( 1 + 3k )² - 8 ( 1 + 3k ) = 0
4 ( 1 + 3k ) [ ( 1 + 3k ) - 2 ] = 0
Therefore ,
1 + 3k = 0 or 1 + 3k - 2 = 0
3k = -1 or 3k = 1
k = -1/3 or k = 1/3
I hope this helps you.
: )
compare x² - 2( 1 + 3k )x + 7( 3 + 2k ) = 0
with ax² + bx + c = 0 we get ,
a = 1 ,
b = - 2 ( 1 + 3k ) ,
c = 7 ( 3 + 2k ) ,
discreaminant = 0
[ since roots are equal ]
b² - 4ac = 0
[ - 2 ( 1 + 3k ) ]² - 4 × 1 × [ 2 ( 1 + 3k ) ] = 0
[2 ( 1 + 3k ]² - 4 [ 2 ( 1 + 3k ) ] = 0
4 ( 1 + 3k )² - 8 ( 1 + 3k ) = 0
4 ( 1 + 3k ) [ ( 1 + 3k ) - 2 ] = 0
Therefore ,
1 + 3k = 0 or 1 + 3k - 2 = 0
3k = -1 or 3k = 1
k = -1/3 or k = 1/3
I hope this helps you.
: )
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