Question

Alguém me ajuda?
2. Calcule k de modo que uma das raízes de 2x^2 – 16x+ 3k + 9 = 0 seja o triplo da outra.

Answer 1

Resposta:

2x² - 16x + 3k + 9 = 0

Pela fórmula de bhaskara:

$\sf x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\sf x=\dfrac{-(-16)\pm\sqrt{(-16)^2-4(2)(3k+9)}}{2(2)}$

$\sf x=\dfrac{16\pm\sqrt{256-8(3k+9)}}{4}\Rightarrow x=\dfrac{16\pm\sqrt{256-24k-72}}{4}$

$\sf x=\dfrac{16\pm\sqrt{256-8(3k+9)}}{4}\Rightarrow x=\dfrac{16\pm\sqrt{184-24k}}{4}$

$\sf x_1=\dfrac{16+\sqrt{184-24k}}{4}~\land~x_2=\dfrac{16-\sqrt{184-24k}}{4}$

• Calcular k de modo que x₁ = 3x₂:

$\sf \dfrac{16+\sqrt{184-24k}}{4}=3\cdot\dfrac{16-\sqrt{184-24k}}{4}$

$\sf \dfrac{16+\sqrt{184-24k}}{4}=\dfrac{48-3\sqrt{184-24k}}{4}$

$\sf 16+\sqrt{184-24k}=48-3\sqrt{184-24k}$

$\sf \sqrt{184-24k}+3\sqrt{184-24k}=48-16$

$\sf 4\sqrt{184-24k}=32$

$\sf \sqrt{184-24k}=\frac{32}{4}$

$\sf \sqrt{184-24k}=8$

$\sf 184-24k=8^2$

$\sf 184-24k=64$

$\sf24k=184-64$

$\sf24k=120$

$\sf k=\frac{120}{24}$

$\red{\sf k=5}$