Question

Find the value of a and b for which x=3/4 and x= -2 are the roots of the equation ax2+bx-6=0.​

Answer 1

Answer:

x = 3/4

x = -2

ax2 + bx - 6 = 0

=> a*3/4^2 + b*-2 = 0

=> 3a/4^2 + -2b = 0

=> 3a/4 + -2b = √0

=> 3a + -2b = 0*4

=> 3a - 2b = 0

=> 3a = 0 + 2b

=> 3a = 2b

=> a/b = 2/3

Therefore,

a = 2. ANS.

b = 3. ans.

Answer 2

$\tt{given : }$

$\tt{ \dag \:x = \frac{3}{4} }$

$\tt{ \dag \: x = - 2}$

the given equation is

$\tt{a {x}^{2} + bx - 6 = 0 }$

so..we have to fill the values of x in the equation.

$\tt{a( \frac{4}{3}) {}^{2} + b( \frac{4}{3} ) - 6 = 0}$

$\tt{a( \frac{16}{9} ) + b( \frac{4}{3} ) - 6 = 0}$

$\tt{ \frac{16a}{9} + \frac{4b}{3} - 6 = 0}$

$\tt{ \frac{16a}{9} + \frac{4b - 18}{3} = 0}$

LCM is equalled to 9

$\tt{\frac{16a}{9} \times 9 + \frac{4b - 18}{3} \times 9 = 0 }$

$\tt{ \frac{16a + 3(4b - 18)}{9} = 0 }$

$\tt{16a + 12b - 54 = 0}$

Divide whole equation by 2

$\tt{8a + 6b - 27 = 0}$