# Question f41ff

Mar 12, 2017

$A + B = 40$

#### Explanation:

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

$A = {\textcolor{b l u e}{9}}^{2} - 4 \left(\textcolor{b l u e}{9}\right) = 81 - 36 = 45$

So $B$ is the other solution to:

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

Add $4$ to both sides to get:

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

That is:

${\left(x - 2\right)}^{2} = {7}^{2}$

Hence:

$x - 2 = \pm 7$

Add $2$ to both sides to get:

$x = 2 \pm 7$

That is:

$x = 9 \text{ }$ or $\text{ } x = - 5$

So $B = - 5$

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

Mar 14, 2017

$A + B + 40.$

#### Explanation:

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

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

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

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

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

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

Enjoy Maths.!