If (x2 + x + 1) + (x2 + 2x + 3) + (x2 + 3x + 5) + ... + (x2 + 20x + 39) = 4500, find the value of x.

Given equation is x2+x+1+x2+2x+3+x2+3x+5+…+x2+20x+39=4500⇒20x2+(1+2+3+…+20)x +(1+3+…+39)=4500⇒ 20x2+20×212x+(20)2=4500⇒ x2+212x+(20)=225⇒ 2x2+21x+40=450⇒ 2x2+21x−410=0⇒ 2x2−20x+41x−410=0⇒ 2x(x−10)+41(x−10)=0⇒ (x−10)(2x+41)=0∴ x=10, -20.5Hence the solution set is {10, -20.5}.

If (x2 + x + 1) + (x2 + 2x + 3) + (x2 + 3x + 5) + ... + (x2 + 20x + 39) = 4500, find the value of x.

Given equation is x2+x+1+x2+2x+3+x2+3x+5+…+x2+20x+39=4500⇒20x2+(1+2+3+…+20)x +(1+3+…+39)=4500⇒ 20x2+20×212x+(20)2=4500⇒ x2+212x+(20)=225⇒ 2x2+21x+40=450⇒ 2x2+21x−410=0⇒ 2x2−20x+41x−410=0⇒ 2x(x−10)+41(x−10)=0⇒ (x−10)(2x+41)=0∴ x=10, -20.5Hence the solution set is {10, -20.5}.

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If (x² + x + 1) + (x² + 2x + 3) + (x² + 3x + 5) + ... + (x² + 20x + 39) = 4500, find the value of x.

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{10, 120.5}

{-10, 20.5}

{10, -20.5}

answer is D.

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Detailed Solution

Given equation is

$\begin{array}{l} (x^{2} + x + 1) + (x^{2} + 2 x + 3) + (x^{2} + 3 x + 5) \\ + \dots + (x^{2} + 20 x + 39) = 4500 \\ \Rightarrow 20 x^{2} + (1 + 2 + 3 + \dots + 20) x \\ + (1 + 3 + \dots + 39) = 4500 \\ \Rightarrow 20 x^{2} + (\frac{20 \times 21}{2}) x + (20)^{2} = 4500 \\ \Rightarrow x^{2} + \frac{21}{2} x + (20) = 225 \\ \Rightarrow 2 x^{2} + 21 x + 40 = 450 \\ \Rightarrow 2 x^{2} + 21 x - 410 = 0 \\ \Rightarrow 2 x^{2} - 20 x + 41 x - 410 = 0 \\ \Rightarrow 2 x (x - 10) + 41 (x - 10) = 0 \\ \Rightarrow (x - 10) (2 x + 41) = 0 \\ ∴ x = 10, - 20.5 \end{array}$