Problem 141
Question
The number of real solutions of the equation \(x^{2}-3\) \(|x|+2=0\) is (A) 2 (B) 4 (C) 1 (D) 3
Step-by-Step Solution
Verified Answer
The equation has 4 real solutions.
1Step 1: Recognize the Absolute Value
The equation given is \( x^2 - 3|x| + 2 = 0 \). The absolute value \(|x|\) suggests that we analyze the equation separately for two cases: \(x \geq 0\) and \(x < 0\).
2Step 2: Case 1 - Positive Values
For \(x \geq 0\), the absolute value \(|x|\) becomes \(x\). Therefore, the equation becomes \(x^2 - 3x + 2 = 0\). We need to solve this quadratic equation.
3Step 3: Solve the Quadratic Equation for Case 1
The equation \(x^2 - 3x + 2 = 0\) factors into \((x - 1)(x - 2) = 0\). This gives the solutions \(x = 1\) and \(x = 2\), which are valid as both are \(\geq 0\).
4Step 4: Case 2 - Negative Values
For \(x < 0\), the absolute value \(|x|\) becomes \(-x\). Therefore, the equation becomes \(x^2 + 3x + 2 = 0\). Solve this quadratic equation.
5Step 5: Solve the Quadratic Equation for Case 2
The equation \(x^2 + 3x + 2 = 0\) factors into \((x + 1)(x + 2) = 0\). This gives the solutions \(x = -1\) and \(x = -2\), which are valid as both are \(< 0\).
6Step 6: Count the Total Solutions
Since both scenarios give us two solutions each that satisfy their respective domain conditions, the equation has a total of four real solutions: \(x = 1, 2, -1, -2\).
Key Concepts
Absolute Value EquationsQuadratic EquationsReal Solutions
Absolute Value Equations
Understanding absolute value is essential when solving equations like this. The absolute value of a number is its distance from zero on a number line, regardless of direction. This means an absolute value expression like \(|x|\) is equal to \(x\) if \(x\) is positive or zero, and \(-x\) if \(x\) is negative.
When dealing with absolute value equations, always consider both positive and negative scenarios. This often results in two different equations. In our example, since we have \( |x| \), we split it into cases:
When dealing with absolute value equations, always consider both positive and negative scenarios. This often results in two different equations. In our example, since we have \( |x| \), we split it into cases:
- For \(x \geq 0\), \(|x|\) simplifies to \(x\).
- For \(x < 0\), \(|x|\) simplifies to \(-x\).
Quadratic Equations
A quadratic equation is a second-degree polynomial equation in the form of \(ax^2 + bx + c = 0\). Solving quadratic equations is a fundamental skill. There are multiple methods to do so, but factoring is often straightforward when the equation is factorable.
In our exercise, after handling the absolute value, we are left with two simpler quadratic equations:
If factorization is challenging, the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) can also be used.
In our exercise, after handling the absolute value, we are left with two simpler quadratic equations:
- For \(x \geq 0\): \(x^2 - 3x + 2 = 0\) which factors to \((x - 1)(x - 2) = 0\), giving solutions \(x = 1\) and \(x = 2\).
- For \(x < 0\): \(x^2 + 3x + 2 = 0\) which factors to \((x + 1)(x + 2) = 0\), yielding solutions \(x = -1\) and \(x = -2\).
If factorization is challenging, the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) can also be used.
Real Solutions
Real solutions refer to the real number roots of an equation. In the context of quadratic and absolute value equations, solutions are the values of \(x\) that satisfy the equation.
It's important to ensure that all solutions conform to the conditions set by any absolute value manipulations applied earlier.
In our exercise, we carefully considered the values of \(x\) in two cases based on the absolute value:
Therefore, the real solutions of the equation \(x^2 - 3|x| + 2 = 0\) are \(x = 1, 2, -1, -2\), confirming there are four real roots in total.
It's important to ensure that all solutions conform to the conditions set by any absolute value manipulations applied earlier.
In our exercise, we carefully considered the values of \(x\) in two cases based on the absolute value:
- For \(x \geq 0\), we found solutions \(x = 1\) and \(x = 2\).
- For \(x < 0\), solutions \(x = -1\) and \(x = -2\) emerged.
Therefore, the real solutions of the equation \(x^2 - 3|x| + 2 = 0\) are \(x = 1, 2, -1, -2\), confirming there are four real roots in total.
Other exercises in this chapter
Problem 138
In the following questions an Assertion (A) is given followed by a Reason \((R) .\) Mark your responses from the following options: (A) Assertion(A) is True and
View solution Problem 139
If \(\alpha \neq \beta\) with \(a^{2}=5 \alpha-3\) and \(\beta^{2}=5 \beta-3\), then \(t h\) equation having \(\alpha^{\prime} \beta\) and \(\beta^{\prime} \alp
View solution Problem 142
The value of ' \(a\) ' for which one root of the quadratic equation \(\left(a^{2}-5 a+3\right) x^{2}+(3 a-1) x+2=0\) is twice as large as the other, is (A) \(\f
View solution Problem 143
If \((1-p)\) is a root of quadratic equation \(x^{2}+p x+\) \((1-p)=0\), then its roots are (A) 0,1 (B) \(-1,2\) (C) \(0,-1\) (D) \(-1,1\)
View solution