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If f(x)=(x2+3x)(x2+3x+2)f(x)=(x^2+3x)(x^2+3x+2), then the sum of all real roots of the equation f(x)+1=9701\sqrt{f(x)+1}=9701, is

Solution

✅ Correct Option: 3

Given f(x)=(x2+3x)(x2+3x+2)f(x) = (x^2+3x)(x^2+3x+2)

Let u=x2+3xu = x^2 + 3x, so:

f(x)=u(u+2)=u2+2uf(x) = u(u+2) = u^2 + 2u


f(x)+1=u2+2u+1=(u+1)2=(x2+3x+1)2f(x) + 1 = u^2 + 2u + 1 = (u+1)^2 = (x^2 + 3x + 1)^2

So the equation f(x)+1=9701\sqrt{f(x)+1} = 9701 becomes:

(x2+3x+1)2=9701\sqrt{(x^2+3x+1)^2} = 9701

∣x2+3x+1∣=9701|x^2+3x+1| = 9701


Case 1: x2+3x+1=9701x^2 + 3x + 1 = 9701

x2+3x−9700=0x^2 + 3x - 9700 = 0

Δ=9+4(9700)=38809=1972>0\Delta = 9 + 4(9700) = 38809 = 197^2 > 0

This gives two real roots.


Case 2: x2+3x+1=−9701x^2 + 3x + 1 = -9701

x2+3x+9702=0x^2 + 3x + 9702 = 0

Δ=9−4(9702)=−38799<0\Delta = 9 - 4(9702) = -38799 < 0

No real roots.


All real roots come from Case 1: x2+3x−9700=0x^2 + 3x - 9700 = 0

Using Vieta's formulas, sum of roots =−ba=−31=−3= -\dfrac{b}{a} = -\dfrac{3}{1} = \boxed{-3}

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