Question

The sum of the abscissae of the points where the curve,
y= kx^2 +(5k+3)x+6k+5, kbelongs to R
Touch the x axis is equal to...
_4/3
-19/3
-10/3
5/3

Answer 1

Good Morning!!!

The graph of this curve is upward parabola and it has two roots.

Both the roots are equal becoz it only touches the x-axis. so, it's Descrimnant = 0

To find absecia put y = 0

kx² + ( 5k + 3 )x + 6k + 5 = 0

it's Descrimnant is

( 5k + 3 )² - 4k ( 6k + 5 ) = 0

=>

25k² + 9 + 30k - 24k² - 20k = 0

=>

k² + 10k + 9 = 0

=>

k² + 9k + 1k + 9 = 0

=>

k ( k + 9 ) + 1 ( k + 9 ) = 0

=>

k = -1 OR k = -9

For k = -1

-x² - 2x - 1 = 0

=>

x² + 2x + 1 = 0

=>

x² + x + x + 1 = 0

=>

x ( x + 1 ) + 1 ( x + 1 ) = 0

x = -1

For x = -9

-9x² -42x -49 = 0

=>

9x² + 42x + 49 = 0

=>

x = (-42 ± √{1764 - 1764}) /18

=>

x = ( -42 )/18

=>

x = -21/9

=>

x = -7/3

So, Sum of absecia is X

X = -1 + ( -7/3 )

=>

X = -10/3