if α and β are the roots of the equation 3x^2 + (x/2) − 4= 0 find p is α−β are the roots of x^2 −px + 7 =0


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Question Number 34184 by Rio Mike last updated on 02/May/18

$i f α a n d β a r e t h e r o o t s o f t h e e q u a t i o n$ $3 x^{2} + \frac{x}{2} - 4 = 0 f i n d p i s α - β a r e$ $t h e r o o t s o f x^{2} - p x + 7 = 0$
Commented by candre last updated on 02/May/18

$3 x^{2} + \frac{x}{2} - 4 = 0$ $6 x^{2} + x - 8 = 0$ $Δ = 1^{2} - 4 \times 6 \times - 8 = 1 + 6 \times 4 \times 8 = 1 + 12 \times 4 \times 4$ $= 1 + 24 \times 2 \times 4 = 1 + 48 \times 2 \times 2 = 1 + 96 \times 2$ $= 1 + 192 = 193$ $x = \frac{- 1 \pm \sqrt{193}}{6}$ $x_{1} = - \frac{1 + \sqrt{193}}{6}; x_{2} = \frac{- 1 + \sqrt{193}}{6}$ $x_{1} + x_{2} = - \frac{1}{3}$ $∣ x_{1} - x_{2} ∣= \frac{\sqrt{193}}{3}$
	Answered by Joel578 last updated on 02/May/18

	$3 x^{2} + \frac{1}{2} x - 4 = 0$ $α - β = ∣ \frac{\sqrt{D}}{a} ∣, where D is discriminant$ $= ∣ \frac{\sqrt{\frac{193}{4}}}{3} ∣ = \frac{\sqrt{193}}{6}$ $x = \frac{\sqrt{193}}{6} is one of the root from equation x^{2} - p x + 7 = 0$ $Let the other root is a$ $(x - \frac{\sqrt{193}}{6}) (x - a) = x^{2} - p x + 7 = 0$ $x^{2} - (a + \frac{\sqrt{193}}{6}) x + \frac{a \sqrt{193}}{6} = x^{2} - p x + 7 = 0$ $\to \frac{a \sqrt{193}}{6} = 7 \to a = \frac{42}{\sqrt{193}}$ $\to p = \frac{42}{\sqrt{193}} + \frac{\sqrt{193}}{6}$
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