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Mathematics

Solve the following quadratic equation:

3x2 + 6x - 4 = 0

Give your answer correct to two places of decimals.

Quadratic Equations

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Answer

Given,

⇒ 3x2 + 6x - 4 = 0

Comparing equation 3x2 + 6x - 4 = 0 with ax2 + bx + c = 0, we get :

a = 3, b = 6 and c = -4.

By formula,

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

Substituting values we get :

⇒x=−(6)±(6)2−4×3×(−4)2×3=−6±36+486=−6±846=−6±21×46=−6±2216=−6+2216 or −6−2216=2(−3+21)6 or 2(−3−21)6=−3+213 or −3−213=−3+4.583 or −3−4.583=1.583 or −7.583=0.53 or −2.53\Rightarrow x = \dfrac{-(6) \pm \sqrt{(6)^2 - 4 \times 3 \times (-4)}}{2 \times 3} \\[1em] = \dfrac{-6 \pm \sqrt{36 + 48}}{6} \\[1em] = \dfrac{-6 \pm \sqrt{84}}{6} \\[1em] = \dfrac{-6 \pm \sqrt{21 \times 4}}{6} \\[1em] = \dfrac{-6 \pm 2\sqrt{21}}{6} \\[1em] = \dfrac{-6 + 2\sqrt{21}}{6} \text{ or } \dfrac{-6 - 2\sqrt{21}}{6} \\[1em] = \dfrac{2(-3 + \sqrt{21})}{6} \text{ or } \dfrac{2(-3 - \sqrt{21})}{6} \\[1em] = \dfrac{-3 + \sqrt{21}}{3} \text{ or } \dfrac{-3 - \sqrt{21}}{3} \\[1em] = \dfrac{-3 + 4.58}{3} \text{ or } \dfrac{-3 - 4.58}{3} \\[1em] = \dfrac{1.58}{3} \text{ or } -\dfrac{7.58}{3} \\[1em] = 0.53 \text{ or } -2.53 \\[1em]

Hence, x = 0.53 or -2.53.

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