The number of distinct real roots of the equation |x||x+2|-5|x+1|-1=0 is \underline\hspace2cm .

Quadratic Equation and Inequalities PYQ — JEE Main 2024 · 5 Apr · Shift 1 · Q60 with Solution

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Correct answer: 3

We need to solve $|x||x+2|-5|x+1|-1=0$ and count the number of distinct real roots.
The absolute values change sign at the points:

\quad x=-1, \quad x=0.$$ So we solve piecewise on the intervals: - $(-\infty,-2)$ - $[-2,-1)$ - $[-1,0)$ - $[0,\infty)$ --- ### Case 1: $x<-2$ Here, $$|x|=-x,\quad |x+2|=-(x+2),\quad |x+1|=-(x+1).$$ So $$|x||x+2|=(-x)(-(x+2))=x(x+2)=x^2+2x.$$ The equation becomes $$x^2+2x-5(-x-1)-1=0$$ $$x^2+2x+5x+5-1=0$$ $$x^2+7x+4=0.$$ Now solve: $$x=\frac{-7\pm\sqrt{49-16}}{2}=\frac{-7\pm\sqrt{33}}{2}.$$ Check which root lies in $x<-2$: - $\frac{-7-\sqrt{33}}{2}<-2$ valid - $\frac{-7+\sqrt{33}}{2}\approx -0.63$ not valid So this case gives **1 root**. --- ### Case 2: $-2\le x<-1$ Here, $$|x|=-x,\quad |x+2|=x+2,\quad |x+1|=-(x+1).$$ Thus $$|x||x+2|=(-x)(x+2)=-x^2-2x.$$ Equation: $$-x^2-2x-5(-x-1)-1=0$$ $$-x^2-2x+5x+5-1=0$$ $$-x^2+3x+4=0$$ $$x^2-3x-4=0$$ $$(x-4)(x+1)=0.$$ Possible roots: $x=4,-1$. In the interval $[-2,-1)$, neither works because: - $x=4$ is outside - $x=-1$ is not included in this interval So this case gives **0 roots**. --- ### Case 3: $-1\le x<0$ Here, $$|x|=-x,\quad |x+2|=x+2,\quad |x+1|=x+1.$$ Thus $$|x||x+2|=(-x)(x+2)=-x^2-2x.$$ Equation: $$-x^2-2x-5(x+1)-1=0$$ $$-x^2-2x-5x-5-1=0$$ $$-x^2-7x-6=0$$ $$x^2+7x+6=0$$ $$(x+1)(x+6)=0.$$ Possible roots: $x=-1,-6$. In the interval $[-1,0)$, only $x=-1$ is valid. So this case gives **1 root**. --- ### Case 4: $x\ge 0$ Here, $$|x|=x,\quad |x+2|=x+2,\quad |x+1|=x+1.$$ So $$|x||x+2|=x(x+2)=x^2+2x.$$ Equation: $$x^2+2x-5(x+1)-1=0$$ $$x^2+2x-5x-5-1=0$$ $$x^2-3x-6=0.$$ Solve: $$x=\frac{3\pm\sqrt{9+24}}{2}=\frac{3\pm\sqrt{33}}{2}.$$ Check $x\ge 0$: - $\frac{3+\sqrt{33}}{2}>0$ valid - $\frac{3-\sqrt{33}}{2}<0$ invalid So this case gives **1 root**. --- 5. Total distinct real roots: $$1+0+1+1=3.$$ Therefore, the number of distinct real roots is $$\boxed{3}.$$

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View written solutionFree

Correct answer: 3

We need to solve $|x||x+2|-5|x+1|-1=0$ and count the number of distinct real roots.
The absolute values change sign at the points:

\quad x=-1, \quad x=0.$$ So we solve piecewise on the intervals: - $(-\infty,-2)$ - $[-2,-1)$ - $[-1,0)$ - $[0,\infty)$ --- ### Case 1: $x<-2$ Here, $$|x|=-x,\quad |x+2|=-(x+2),\quad |x+1|=-(x+1).$$ So $$|x||x+2|=(-x)(-(x+2))=x(x+2)=x^2+2x.$$ The equation becomes $$x^2+2x-5(-x-1)-1=0$$ $$x^2+2x+5x+5-1=0$$ $$x^2+7x+4=0.$$ Now solve: $$x=\frac{-7\pm\sqrt{49-16}}{2}=\frac{-7\pm\sqrt{33}}{2}.$$ Check which root lies in $x<-2$: - $\frac{-7-\sqrt{33}}{2}<-2$ valid - $\frac{-7+\sqrt{33}}{2}\approx -0.63$ not valid So this case gives **1 root**. --- ### Case 2: $-2\le x<-1$ Here, $$|x|=-x,\quad |x+2|=x+2,\quad |x+1|=-(x+1).$$ Thus $$|x||x+2|=(-x)(x+2)=-x^2-2x.$$ Equation: $$-x^2-2x-5(-x-1)-1=0$$ $$-x^2-2x+5x+5-1=0$$ $$-x^2+3x+4=0$$ $$x^2-3x-4=0$$ $$(x-4)(x+1)=0.$$ Possible roots: $x=4,-1$. In the interval $[-2,-1)$, neither works because: - $x=4$ is outside - $x=-1$ is not included in this interval So this case gives **0 roots**. --- ### Case 3: $-1\le x<0$ Here, $$|x|=-x,\quad |x+2|=x+2,\quad |x+1|=x+1.$$ Thus $$|x||x+2|=(-x)(x+2)=-x^2-2x.$$ Equation: $$-x^2-2x-5(x+1)-1=0$$ $$-x^2-2x-5x-5-1=0$$ $$-x^2-7x-6=0$$ $$x^2+7x+6=0$$ $$(x+1)(x+6)=0.$$ Possible roots: $x=-1,-6$. In the interval $[-1,0)$, only $x=-1$ is valid. So this case gives **1 root**. --- ### Case 4: $x\ge 0$ Here, $$|x|=x,\quad |x+2|=x+2,\quad |x+1|=x+1.$$ So $$|x||x+2|=x(x+2)=x^2+2x.$$ Equation: $$x^2+2x-5(x+1)-1=0$$ $$x^2+2x-5x-5-1=0$$ $$x^2-3x-6=0.$$ Solve: $$x=\frac{3\pm\sqrt{9+24}}{2}=\frac{3\pm\sqrt{33}}{2}.$$ Check $x\ge 0$: - $\frac{3+\sqrt{33}}{2}>0$ valid - $\frac{3-\sqrt{33}}{2}<0$ invalid So this case gives **1 root**. --- 5. Total distinct real roots: $$1+0+1+1=3.$$ Therefore, the number of distinct real roots is $$\boxed{3}.$$

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