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Mathematics

Solve the following equation using the formula :

23x=−16x2−13\dfrac{2}{3}x = -\dfrac{1}{6}x^2 - \dfrac{1}{3}

Quadratic Equations

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Answer

Given,

⇒23x=−16x2−13⇒23x=−x2−26⇒12x=3(−x2−2)⇒12x=−3x2−6⇒3x2+12x+6=0.\Rightarrow \dfrac{2}{3}x = -\dfrac{1}{6}x^2 - \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{2}{3}x = \dfrac{-x^2 - 2}{6} \\[1em] \Rightarrow 12x = 3(-x^2 - 2) \\[1em] \Rightarrow 12x = -3x^2 - 6 \\[1em] \Rightarrow 3x^2 + 12x + 6 = 0.

Comparing 3x2 + 12x + 6 = 0 with ax2 + bx + c = 0 we get,

a = 3, b = 12, c = 6.

We know that,

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting values we get,

⇒x=−(12)±(12)2−4(3)(6)2(3)=−12±144−726=−12±726=−12±626=−2±2=−2+2 or −2−2=−0.59 or −3.41\Rightarrow x = \dfrac{-(12) \pm \sqrt{(12)^2 - 4(3)(6)}}{2(3)} \\[1em] = \dfrac{-12 \pm \sqrt{144 - 72}}{6} \\[1em] = \dfrac{-12 \pm \sqrt{72}}{6} \\[1em] = \dfrac{-12 \pm 6\sqrt{2}}{6} \\[1em] = -2 \pm \sqrt{2} \\[1em] = -2 + \sqrt{2} \text{ or } -2 - \sqrt{2} \\[1em] = -0.59 \text{ or } -3.41

Hence, x = -0.59, -3.41.

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