Solving the Absolute Value Equation
We need to find the number of real roots for the equation $x |x-2|+3|x-3|+1=0$. We will solve this by considering different cases based on the values of $x$ where the expressions inside the absolute value signs change sign.
The critical points are $x=2$ and $x=3$. These points divide the number line into three intervals: $x < 2$, $2 \le x < 3$, and $x \ge 3$. We analyze the equation in each interval.
Case Analysis
Case 1: $x < 2$
- In this interval, $|x-2| = -(x-2) = 2-x$ and $|x-3| = -(x-3) = 3-x$.
- The equation becomes: $x(2-x) + 3(3-x) + 1 = 0$.
- Simplifying: $2x - x^2 + 9 - 3x + 1 = 0$.
- This gives the quadratic equation: $-x^2 - x + 10 = 0$, or $x^2 + x - 10 = 0$.
- Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-10)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 40}}{2} = \frac{-1 \pm \sqrt{41}}{2}$.
- The two potential roots are $x_1 = \frac{-1 + \sqrt{41}}{2}$ and $x_2 = \frac{-1 - \sqrt{41}}{2}$.
- We check if these roots satisfy the condition $x < 2$.
- $\sqrt{41}$ is between 6 and 7 (approx 6.4).
- $x_1 \approx \frac{-1 + 6.4}{2} \approx 2.7$. This is not less than 2.
- $x_2 \approx \frac{-1 - 6.4}{2} \approx -3.7$. This is less than 2.
- Therefore, $x = \frac{-1 - \sqrt{41}}{2}$ is a valid root.
Case 2: $2 \le x < 3$
- In this interval, $|x-2| = x-2$ and $|x-3| = -(x-3) = 3-x$.
- The equation becomes: $x(x-2) + 3(3-x) + 1 = 0$.
- Simplifying: $x^2 - 2x + 9 - 3x + 1 = 0$.
- This gives the quadratic equation: $x^2 - 5x + 10 = 0$.
- Calculate the discriminant $\Delta = b^2 - 4ac = (-5)^2 - 4(1)(10) = 25 - 40 = -15$.
- Since the discriminant $\Delta < 0$, there are no real roots in this interval.
Case 3: $x \ge 3$
- In this interval, $|x-2| = x-2$ and $|x-3| = x-3$.
- The equation becomes: $x(x-2) + 3(x-3) + 1 = 0$.
- Simplifying: $x^2 - 2x + 3x - 9 + 1 = 0$.
- This gives the quadratic equation: $x^2 + x - 8 = 0$.
- Using the quadratic formula: $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-8)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 32}}{2} = \frac{-1 \pm \sqrt{33}}{2}$.
- The two potential roots are $x_3 = \frac{-1 + \sqrt{33}}{2}$ and $x_4 = \frac{-1 - \sqrt{33}}{2}$.
- We check if these roots satisfy the condition $x \ge 3$.
- $\sqrt{33}$ is between 5 and 6 (approx 5.7).
- $x_3 \approx \frac{-1 + 5.7}{2} \approx 2.35$. This is not greater than or equal to 3.
- $x_4 \approx \frac{-1 - 5.7}{2} \approx -3.35$. This is not greater than or equal to 3.
- Therefore, there are no valid roots in this interval.
Conclusion
Combining the results from all three cases, we found only one real root: $x = \frac{-1 - \sqrt{41}}{2}$.
The number of real roots is 1.
Similar Questions
Let A = {-3, -2, -1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if $0\le x^2+2y\le 4$. Let $l$ be the number of elements in R and $m$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $l + m$ is equal to
Let $\alpha$ and $\beta$ be the roots of $x^2 + \sqrt{3}x-16=0$, and $\gamma$ and $\delta$ be the roots of $x^2 + 3x - 1 = 0$. If $P_n = \alpha^n + \beta^n$ and $Q_n = \gamma^n + \delta^n$, then $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to
Let the domain of the function $f (x) = \log_2 \log_4 \log_6(3 + 4x -x^2)$ be $(a, b)$.
If $\int_{b-a}^{b+a}[x^2] dx=p-\sqrt{q}-\sqrt{r}$, $p,q,r\in N$, $gcd(p,q,r) = 1$, where $[.]$ is the greatest integer function, then $p + q + r$ is equal to
Let A be a matrix of order $3 \times 3$ and $|A| = 5$. If $|2\text{adj} (3A \text{adj} (2A))| = 2^\alpha \cdot 3^\beta \cdot 5^\gamma$, $\alpha, \beta, \gamma \in N$, then $\alpha + \beta + \gamma$ is equal to
Let $a_1, a_2, a_3,....$ be a G.P. of increasing positive numbers. If $a_3a_5 = 729$ and $a_2 + a_4 = \frac{111}{4}$, then $24 (a_1 + a_2 + a_3)$ is equal to
All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number $n$ be denoted by $W_n$. Let the probability $P(W_n)$ of choosing the word $W_n$ satisfy $P(W_n) = 2P(W_{n-1})$, $n > 1$.
If $P(CDBEA) = \frac{2^\alpha}{2^\beta-1}$, $\alpha, \beta\in N$, then $\alpha + \beta$ is equal to :
- The largest $n \in N$ such that $3^n$ divides $50!$ is :
- The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five 1s and exactly three 2s, is equal to :
- Let A be the set of all functions $f: Z \to Z$ and R be a relation on A such that $R = \{(f, g) : f(0) = g(1) \text{ and } f(1) = g(0)\}$. Then R is :
Let $a \in \mathbf{R}$ and $A$ be a matrix of order $3 \times 3$ such that $\det (A) = -4$ and $A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}$, where $I$ is the identity matrix of order $3 \times 3$. If $\det ((a+1)\text{adj}((a-1)A))$ is $2^m 3^n$, $m, n \in \{0, 1, 2, \dots, 20\}$, then $m+n$ is equal to :
Important Questions from Algebra
Let A = {-3, -2, -1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if $0\le x^2+2y\le 4$. Let $l$ be the number of elements in R and $m$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $l + m$ is equal to
Let $\alpha$ and $\beta$ be the roots of $x^2 + \sqrt{3}x-16=0$, and $\gamma$ and $\delta$ be the roots of $x^2 + 3x - 1 = 0$. If $P_n = \alpha^n + \beta^n$ and $Q_n = \gamma^n + \delta^n$, then $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to
Let the domain of the function $f (x) = \log_2 \log_4 \log_6(3 + 4x -x^2)$ be $(a, b)$.
If $\int_{b-a}^{b+a}[x^2] dx=p-\sqrt{q}-\sqrt{r}$, $p,q,r\in N$, $gcd(p,q,r) = 1$, where $[.]$ is the greatest integer function, then $p + q + r$ is equal to
Let A be a matrix of order $3 \times 3$ and $|A| = 5$. If $|2\text{adj} (3A \text{adj} (2A))| = 2^\alpha \cdot 3^\beta \cdot 5^\gamma$, $\alpha, \beta, \gamma \in N$, then $\alpha + \beta + \gamma$ is equal to
Let $a_1, a_2, a_3,....$ be a G.P. of increasing positive numbers. If $a_3a_5 = 729$ and $a_2 + a_4 = \frac{111}{4}$, then $24 (a_1 + a_2 + a_3)$ is equal to
