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Answer 1 · 2017-03-12T00:23:39

$A+B=40$

If $9$ is a solution to the equation $x^2-4x=A$ then:

$A = color(blue)(9)^2-4(color(blue)(9)) = 81-36 = 45$

So $B$ is the other solution to:

$x^2-4x=45$

Add $4$ to both sides to get:

$x^2-4x+4 = 49$

That is:

$(x-2)^2 = 7^2$

Hence:

$x-2 = +-7$

Add $2$ to both sides to get:

$x = 2+-7$

That is:

$x = 9" "$ or $" "x = -5$

So $B=-5$

So $A+B = 45+(-5) = 40$

Answer 2 · 2017-03-14T14:22:49

$A+B+40.$

By what is given, $B and 9$ are the roots of the Quadr. Eqn.,

$x^2-4x-A=0.$

Knowing that, in the Quadr. Eqn. $:ax^2+bx+c=0,$

Sum of Roots $=-b/a," and, Product of Roots ="c/a$ , we have,

$B+9=-(-4)/1=4 rArr B=-5, &, 9B=-A/1=-A, or, A=-9B$

Hence, $A+B=-9B+B=-8B=(-8)(-5)=40,$ as Respected George C. Sir has already derived!

Enjoy Maths.!