Problem 79
Question
If the roots of the equation \(x^{2}-2 a x+a^{2}+a-3=0\) are real less than 3 , then: (A) \(a<2\) (B) \(2 \leq a \leq 3\) (C) \(34\)
Step-by-Step Solution
Verified Answer
The correct answer is (B) \(2 \leq a \leq 3\).
1Step 1: Check the Condition for Real Roots
To ensure that the roots of the quadratic equation \(x^2 - 2ax + a^2 + a - 3 = 0\) are real, the discriminant should be non-negative. The discriminant \(\Delta\) for a quadratic equation \(ax^2 + bx + c = 0\) is given by \(\Delta = b^2 - 4ac\). Here, \(a=1\), \(b=-2a\), and \(c=a^2 + a - 3\). So, the discriminant is \((-2a)^2 - 4(1)(a^2 + a - 3)\).
2Step 2: Simplify the Discriminant Inequality
The discriminant expression is simplified as \((4a^2) - 4(a^2 + a - 3)\). Distribute the \(-4\) to get \(4a^2 - 4a^2 - 4a + 12\). Further simplifying gives us \(-4a + 12 \geq 0\). Solving for \(a\) results in \(a \leq 3\).
3Step 3: Apply Condition to Roots Less Than 3
Since the roots of the equation need to be less than 3, by substituting \(x = 3\) into the equation \(x^2 - 2ax + a^2 + a - 3 = 0\), it should be negative: \(9 - 6a + a^2 + a - 3 < 0\). This simplifies to \(a^2 - 5a + 6 < 0\).
4Step 4: Solve the New Quadratic Inequality
Factor the quadratic inequality \(a^2 - 5a + 6 < 0\). It factors to \((a - 2)(a - 3) < 0\). This inequality holds true when \(2 < a < 3\). Therefore, the values of \(a\) for which the original roots are real and less than 3 are within this interval.
Key Concepts
DiscriminantReal RootsInequalitiesQuadratic Inequality Solution
Discriminant
The discriminant is a key component in determining the nature of the roots of a quadratic equation. When a quadratic equation is in the form \(ax^2 + bx + c = 0\), the discriminant \(\Delta\) can be calculated using \(\Delta = b^2 - 4ac\). This single value tells us much about the roots:
- If \(\Delta > 0\), the equation has two distinct real roots.
- If \(\Delta = 0\), the equation has exactly one real root, which is repeated (also known as a double root).
- If \(\Delta < 0\), the equation has no real roots but two complex conjugate roots.
Real Roots
Real roots of a quadratic equation refer to the solutions that are actual numbers, as opposed to complex or imaginary numbers. For real roots to exist, the discriminant discussed earlier must be non-negative.
In the context of our example, which aims for real roots, the task is to evaluate the condition \(-4a + 12 \geq 0\). Solving this inequality gives the condition \(a \leq 3\), ensuring that the quadratic equation will indeed have real roots.
When working with inequalities to find the range of parameter values that provide real roots, always keep in mind that reality reflects tangible or visualizable outcomes in terms of number line or real plane.
In the context of our example, which aims for real roots, the task is to evaluate the condition \(-4a + 12 \geq 0\). Solving this inequality gives the condition \(a \leq 3\), ensuring that the quadratic equation will indeed have real roots.
When working with inequalities to find the range of parameter values that provide real roots, always keep in mind that reality reflects tangible or visualizable outcomes in terms of number line or real plane.
Inequalities
Inequalities are mathematical expressions involving the symbols \(<, >, \leq, \geq\). They express a range or set of values that satisfy a particular condition, instead of one specific value.
In the context of quadratic equations, inequalities help pinpoint the ranges of variable values that yield certain properties for the equation's roots. For instance, ensuring roots are less than a given value involves setting up an inequality. In our problem, we require that the roots of the equation be less than 3, which translates into solving the inequality involving \(9 - 6a + a^2 + a - 3 < 0\).
In the context of quadratic equations, inequalities help pinpoint the ranges of variable values that yield certain properties for the equation's roots. For instance, ensuring roots are less than a given value involves setting up an inequality. In our problem, we require that the roots of the equation be less than 3, which translates into solving the inequality involving \(9 - 6a + a^2 + a - 3 < 0\).
- Factor the quadratic to derive intervals of solution.
- Test intervals to determine which satisfy the original condition.
Quadratic Inequality Solution
Solving a quadratic inequality involves more steps than a standard quadratic equation. Here's a streamlined approach to tackle quadratic inequalities like \(a^2 - 5a + 6 < 0\).
- Start by factorizing the quadratic expression if possible. The expression \(a^2 - 5a + 6\) factors to \((a - 2)(a - 3)\).
- Identify critical points from the factored form: in our case, these are \(a = 2\) and \(a = 3\).
- Create a number line and plot the critical points to divide the line into intervals.
- Determine the sign of each interval by testing points from each section. For \((a - 2)(a - 3) < 0\), the interval \(2 < a < 3\) will satisfy the inequality.
Other exercises in this chapter
Problem 75
The solution set of \((x)^{2}+(x+1)^{2}=25\), where \((x)\) is the least integer greater than or equal to \(x\), is (A) \((2,4)\) (B) \((-5,4] \cup(2,3]\) (C) \
View solution Problem 77
Solution of \(2^{x}+2^{|x|} \geq 2 \sqrt{2}\) is (A) \(\left(-\infty, \log _{2}(\sqrt{2}+1)\right.\) (B) \((0,8)\) (C) \(\left(\frac{1}{2}, \log _{2}(\sqrt{2}-1
View solution Problem 80
The solution set of \(\left|\frac{x+1}{x}\right|+|x+1|=\frac{(x+1)^{2}}{|x|}\) is (A) \(\\{x \mid x \geq 0\\}\) (B) \(\\{x \mid x>0\\} \cup\\{-1\\}\) (C) \(\\{-
View solution Problem 81
If \(\alpha, \beta\) are the roots of the equation \(a x^{2}+b x+c=0\), \((a \neq 0)\) and \(\alpha+\delta, \beta+\delta\) are the roots of \(A x^{2}+B x+\) \(C
View solution