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Mathematics

Use formula to solve the quadratic equation :

x2 + x - (a + 1)(a + 2) = 0

Quadratic Equations

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Answer

Given,

x2 + x - (a + 1)(a + 2) = 0

Comparing above equation with ax2 + bx + c = 0, we get :

a = 1, b = 1 and c = -(a + 1)(a + 2)

By formula,

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

Substituting values we get :

⇒x=−1±12−4×1×−(a+1)(a+2)2×1⇒x=−1±1+4(a+1)(a+2)2⇒x=−1±1+4(a2+2a+a+2)2⇒x=−1±1+4(a2+3a+2)2⇒x=−1±1+4a2+12a+82⇒x=−1±4a2+12a+92⇒x=−1±(2a+3)22⇒x=−1±(2a+3)2⇒x=−1+(2a+3)2,−1−(2a+3)2⇒x=2a+3−12,−1−2a−32⇒x=2a+22,−2a−42⇒x=2(a+1)2,−2(a+2)2⇒x=(a+1),−(a+2).\Rightarrow x = \dfrac{-1 \pm \sqrt{1^2 - 4 \times 1 \times -(a + 1)(a + 2)}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{-1 \pm \sqrt{1 + 4(a + 1)(a + 2)}}{2} \\[1em] \Rightarrow x = \dfrac{-1 \pm \sqrt{1 + 4(a^2 + 2a + a + 2)}}{2} \\[1em] \Rightarrow x = \dfrac{-1 \pm \sqrt{1 + 4(a^2 + 3a + 2)}}{2} \\[1em] \Rightarrow x = \dfrac{-1 \pm \sqrt{1 + 4a^2 + 12a + 8}}{2} \\[1em] \Rightarrow x = \dfrac{-1 \pm \sqrt{4a^2 + 12a + 9}}{2} \\[1em] \Rightarrow x = \dfrac{-1 \pm \sqrt{(2a + 3)^2}}{2} \\[1em] \Rightarrow x = \dfrac{-1 \pm (2a + 3)}{2} \\[1em] \Rightarrow x = \dfrac{-1 + (2a + 3)}{2}, \dfrac{-1 - (2a + 3)}{2} \\[1em] \Rightarrow x = \dfrac{2a + 3 - 1}{2}, \dfrac{-1 - 2a - 3}{2} \\[1em] \Rightarrow x = \dfrac{2a + 2}{2}, \dfrac{-2a - 4}{2} \\[1em] \Rightarrow x = \dfrac{2(a + 1)}{2}, \dfrac{-2(a + 2)}{2} \\[1em] \Rightarrow x = (a + 1), -(a + 2).

Hence, x = (a + 1) or -(a + 2).

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