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 :
Quadratic Equation Roots Analysis
We analyze relationships between roots of quadratic equations to find the value of parameter $m$.
Solving for α, β, and r
Equation 1: $x2 - 3x + r = 0$
- Roots: $α, β$
- Vieta's Sum: $α + β = 3$ (1)
- Vieta's Product: $αβ = r$ (2)
Equation 2: $x2 + 3x + r = 0$
- Roots: $α/2, 2β$
- Vieta's Sum: $α/2 + 2β = -3$ (3)
- Vieta's Product: $(α/2)(2β) = αβ = r$ (4)
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): $r = αβ = (6)(-3) = -18$
Third Equation Roots and Parameter m
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 $R2$: $R2 = 6 - 2(-3) - (-18)/2 = 6 + 6 - (-9) = 12 + 9 = 21$
Determining the Value of m
For Equation 3 ($x2 + 6x - m = 0$), Vieta's formulas state:
- Sum of roots: $R1 + R2 = -6$. (Check: $-27 + 21 = -6$, consistent.)
- Product of roots: $R1R2 = -m$
Calculate the product of the roots:
$R1R2 = (-27)(21) = -567$
Equate this product to $-m$:
$-m = -567$
Therefore, $m = 567$.
Similar Questions
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- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
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 ________.
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