Question

if (2x minus 1) is exactly divisible by 6 X square + ax - 4 and b x square - 11x + 3, then the value of b square minus a square + a minus b is​

Answer 1

Step-by-step explanation:

Given that:-

$(2x - 1)$

is exactly divisible by

$6 {x}^{2} + ax - 4$

and

$b {x}^{2} - 11x + 3$

We need to simply, place the value of x.

$2x - 1 = 0$

$2x = 1$

$x = \frac{1}{2}$

Now,

$6 {( \frac{1}{2} )}^{2} + \frac{a}{2} - 4 = 0$

$\frac{3 + a}{2} = 4$

$3 + a = 4 \times 2 = 8$

$a = 8 - 3$

$a = 5$

Now, to find b.

$b {( \frac{1}{2}) }^{2} - \frac{11}{2} + 3 = 0$

$\frac{b}{4} - \frac{11}{2} = - 3$

$\frac{b - 22}{4} = - 3$

$b - 22 = - 12$

$b = 10$

is the answer.

Now, we need

${b}^{2} - {a}^{2} + a - b$

$100 - 25 + 100 + 5$

$= 205 - 25$

$= 180$

is the answer.

Hope it helps :D