If the roots of the equation $x^2 + 6x = m$ are $2\alpha + \beta + 2r$ and $\alpha - 2\beta - \frac{r}{2}$, then $m$ is equal to :
We analyze relationships between roots of quadratic equations to find the value of parameter $m$.
Equation 1: $x2 - 3x + r = 0$
Equation 2: $x2 + 3x + r = 0$
From (1), we have $α = 3 - β$. Substituting this into (3):
$(3 - β)/2 + 2β = -3$
Multiply by 2 to clear the fraction: $3 - β + 4β = -6$
Simplify: $3 + 3β = -6$
Solve for $β$: $3β = -9$ &implies $β = -3$
Substitute $β = -3$ back into (1): $α + (-3) = 3$ &implies $α = 6$
Calculate $r$ using (2): $r = αβ = (6)(-3) = -18$
Equation 3: $x2 + 6x = m$, rewritten as $x2 + 6x - m = 0$.
The roots are given as $R1 = 2α + β + 2r$ and $R2 = α - 2β - r/2$.
Substitute the found values: $α=6$, $β=-3$, $r=-18$.
Calculate $R1$: $R1 = 2(6) + (-3) + 2(-18) = 12 - 3 - 36 = -27$
Calculate $R2$: $R2 = 6 - 2(-3) - (-18)/2 = 6 + 6 - (-9) = 12 + 9 = 21$
For Equation 3 ($x2 + 6x - m = 0$), Vieta's formulas state:
Calculate the product of the roots:
$R1R2 = (-27)(21) = -567$
Equate this product to $-m$:
$-m = -567$
Therefore, $m = 567$.
Let f and g be functions satisfying $f(x+y) = f(x)f(y), f(1) = 7$ and $g(x+y) = g(xy), g(1) = 1$, for all $x, y \in \mathbb{N}$. If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$, then n is equal to :
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.