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Quadratic Equation and Inequalities question

2017 · 9 Apr · Shift 1 · Q29
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  5. /2017 · 9 Apr · Shift 1 · Q29

Quadratic Equation and Inequalities question

2017 · 9 Apr · Shift 1 · Q29

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
The sum of all the real values of x satisfying the equation 2(x −-− 1)(x2 + 5x −-− 50) = 1 is :
  1. A
    16
  2. B
    14
  3. C
    −-− 4
  4. D
    −-− 5
View written solutionFree

Correct answer: C

  1. Given equation

We need to solve 2(x−1)(x2+5x−50)=1.2(x-1)(x^2+5x-50)=1.2(x−1)(x2+5x−50)=1.

We are asked for the sum of all real values of xxx satisfying it.

  1. Factor the quadratic

First factor: x2+5x−50=(x+10)(x−5).x^2+5x-50=(x+10)(x-5).x2+5x−50=(x+10)(x−5).

So the equation becomes 2(x−1)(x+10)(x−5)=1.2(x-1)(x+10)(x-5)=1.2(x−1)(x+10)(x−5)=1.

  1. Look for a useful substitution

Notice that the three factors are centered around x=1x=1x=1:

  • x−1x-1x−1
  • x+10=(x−1)+11x+10=(x-1)+11x+10=(x−1)+11
  • x−5=(x−1)−4x-5=(x-1)-4x−5=(x−1)−4

Let y=x−1.y=x-1.y=x−1. Then

eq x-5=y-4.$$ So the equation becomes $$2y(y+11)(y-4)=1.$$ Now expand: $$(y+11)(y-4)=y^2+7y-44,$$ so $$2y(y^2+7y-44)=1,$$ $$2y^3+14y^2-88y-1=0.$$ This cubic is not immediately pleasant to solve directly, so let us instead expand in $x$ and inspect its structure. --- 4. **Expand the original equation** We have $$(x-1)(x^2+5x-50)=x^3+4x^2-55x+50.$$ Thus $$2x^3+8x^2-110x+100=1,$$ $$2x^3+8x^2-110x+99=0.$$ So the real roots of the equation are the real roots of $$2x^3+8x^2-110x+99=0.$$ --- 5. **Use Vieta's relation carefully** Let the roots be $r_1,r_2,r_3$. Then $$r_1+r_2+r_3=-\frac{8}{2}=-4.$$ But this is the sum of **all roots**, real or complex. We must check how many roots are real. --- 6. **Check number of real roots** Consider $$f(x)=2x^3+8x^2-110x+99.$$ Evaluate at a few points: $$f(-10)=-2000+800+1100+99=-1<0,$$ $$f(-9)=-1458+648+990+99=279>0.$$ So there is a root in $(-10,-9)$. Next, $$f(0)=99>0, eq f(1)=2+8-110+99=-1<0.$$ So there is a root in $(0,1)$. Next, $$f(5)=250+200-550+99=-1<0,$$ $$f(6)=432+288-660+99=159>0.$$ So there is a root in $(5,6)$. Hence the cubic has **three real roots**. Therefore, the sum of all real values of $x$ equals the sum of all three roots: $$-4.$$ --- 7. **Match with options** The required sum is $$\boxed{-4}.$$ So the correct option is **C**. --- 8. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** They agree.
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