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Chapter 4: Problem 141

The number of real solutions of the equation \(x^{2}-3\) \(|x|+2=0\) is (A) 2 (B) 4 (C) 1 (D) 3

Short Answer

Expert verified
The equation has 4 real solutions.

Step by step solution

01

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 \(x < 0\).
02

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.
03

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 \(\geq 0\).
04

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.
05

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 \(< 0\).
06

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: \(x = 1, 2, -1, -2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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:

  • For \(x \geq 0\), \(|x|\) simplifies to \(x\).
  • For \(x < 0\), \(|x|\) simplifies to \(-x\).
This approach ensures we account for all possible values of \(x\) and is crucial for finding all potential solutions.
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:

  • 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 \(x = -2\).
Factorization involves expressing the quadratic equation as a product of its linear factors, making it easier to find the roots by setting each factor equal to zero.

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:
  • For \(x \geq 0\), we found solutions \(x = 1\) and \(x = 2\).
  • For \(x < 0\), solutions \(x = -1\) and \(x = -2\) emerged.
Each solution was checked against the condition used to define the behavior of \(|x|\).

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.

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Most popular questions from this chapter

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) \(-1,1\)

Given \(l x^{2}-m x+5=0\) does not have two distinct real roots, the minimum value of \(5 l+m\) is (A) 5 (B) \(-5\) (C) 1 (D) \(-1\)

The maximum number of positive real roots of a polynomial equation \(f(x)=0\) is the number of changes of signs from positive to negative and negative to positive in \(f(x)\). For example, consider the equation \(f(x)=x^{3}+6 x^{2}+11 x-\) \(6=0\). The signs of the various terms are: $$ +++- $$ Clearly, there is only one change of sign in the given expression. So, the given equation has at most one positive Ireal root. The maximum number of negative real roots of a polynomial equation \(f(x)=0\) is the number of changes of signs from positive to negative and negative to positive in \(f(-x)\). For example, for the equation \(f(x)=x^{4}+x^{3}+x^{2}-x-1=\) 0 , there are three changes of signs in \(f(-x)\). So, the given equation has atmost three negative real roots. If \(f(x)\) and \(f(-x)\) do not have any changes of signs, the equation \(f(x)=0\) has no real roots. Now, consider the polynomial $$ P_{n}(x)=1+2 x+3 x^{2}+\ldots+(n+1) x^{n} . $$ If \(n\) is even, the number of real roots of \(P_{n}(x)\) is (A) 0 (B) \(n\) (C) 1 (D) None of these

Let \(\alpha\) and \(\beta\) be the roots of equation \(x^{2}-6 x-2=0\). If \(a_{n}=\alpha^{n}-\beta^{n}\), for \(n \geq 1\),then the value of \(\frac{a_{10}-2 a_{8}}{2 a_{9}}\) is equal to (A) \(-6\) (B) 3 (C) \(-3\) (D) 6

If \(f(x)=x-[x], x(\neq 0) \in R\), where \([x]\) is the greatest integer less than or equal to \(x\), then the number of solutions of \(f(x)+f\left(\frac{1}{x}\right)=1\) are (A) 0 (B) 1 (C) infinite (D) 2
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