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Answer 1 · 2017-03-12T00:23:39

$A + B = 40$ $A+B=40$

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

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

So $B$ $B$ is the other solution to:

$x^{2} - 4 x = 45$ $x^2-4x=45$

Add $4$ $4$ to both sides to get:

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

That is:

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

Hence:

$x - 2 = \pm 7$ $x-2 = +-7$

Add $2$ $2$ to both sides to get:

$x = 2 \pm 7$ $x = 2+-7$

That is:

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

So $B = - 5$ $B=-5$

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

Answer 2 · 2017-03-14T14:22:49

$A + B + 40 .$ $A+B+40.$

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

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

Knowing that, in the Quadr. Eqn. $: a x^{2} + b x + c = 0,$ $:ax^2+bx+c=0,$

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

$B + 9 = - \frac{- 4}{1} = 4 \Rightarrow B = - 5, &, 9 B = - \frac{A}{1} = - A, or, A = - 9 B$ $B+9=-(-4)/1=4 rArr B=-5, &, 9B=-A/1=-A, or, A=-9B$

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

Enjoy Maths.!