Problem 17
Question
Both the roots of the equation \((x-b)(x-c)+(x-a)\) \((x-c)+(x-a)(x-b)=0\) are always (A) positive (B) negative (C) real (D) None of these
Step-by-Step Solution
Verified Answer
The roots are always real (Option C).
1Step 1: Expand the Equation
The given equation is \((x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0\). Expand each term separately:1. \((x-b)(x-c) = x^2 - (b+c)x + bc\)2. \((x-a)(x-c) = x^2 - (a+c)x + ac\)3. \((x-a)(x-b) = x^2 - (a+b)x + ab\).Add these expanded terms together.
2Step 2: Combine Like Terms
Combine all the expanded terms from Step 1:\[ x^2 - (b+c)x + bc + x^2 - (a+c)x + ac + x^2 - (a+b)x + ab = 0 \].This simplifies to:\[ 3x^2 - [2(a+b+c)]x + [ab+bc+ac] = 0 \].This is a quadratic equation in the form \(ax^2 + bx + c = 0\).
3Step 3: Use the Discriminant to Check Root Condition
To determine the nature of the roots, compute the discriminant \(D\) of the quadratic equation, where \( D = b^2 - 4ac \). Here:- \(a = 3\),- \(b = -2(a+b+c)\),- \(c = ab+bc+ac\).Calculate \( D \):\[ D = [2(a+b+c)]^2 - 4 \times 3 \times (ab+bc+ac) \]This simplifies to: \[ 4(a+b+c)^2 - 12(ab+bc+ac) \].Since this discriminant involves squares and products, it represents real values ensuring real roots.
4Step 4: Conclusion Based on the Discriminant
Since the quadratic equation always yields a positive discriminant, the roots of the equation are always real. Thus, the correct choice is option (C) real.
Key Concepts
DiscriminantNature of RootsQuadratic Formula
Discriminant
In quadratic equations, the discriminant is a crucial component for understanding the nature of the roots. If you have an equation in the standard form \(ax^2 + bx + c = 0\), the discriminant \(D\) is calculated using the formula \(D = b^2 - 4ac\). This value tells us if the roots are real or complex, and if they are equal or distinct.
- If \(D > 0\), the equation has two distinct real roots.
- If \(D = 0\), there is exactly one real root, also known as a repeated or double root.
- If \(D < 0\), the roots are complex and come as a pair of conjugates.
Nature of Roots
The term 'nature of roots' refers to the type and number of solutions that a quadratic equation possesses. Understanding the nature of roots is pivotal for both theoretical and practical applications of quadratic equations.
In the context of the exercise, the expanded quadratic equation is \(3x^2 - [2(a+b+c)]x + [ab+bc+ac] = 0\). To determine if the roots are real, equal, or complex, we analyze the discriminant. As explained previously, a positive discriminant means that the roots are real, which is confirmed by the expanded discriminant formula \[4(a+b+c)^2 - 12(ab+bc+ac)\].
In the context of the exercise, the expanded quadratic equation is \(3x^2 - [2(a+b+c)]x + [ab+bc+ac] = 0\). To determine if the roots are real, equal, or complex, we analyze the discriminant. As explained previously, a positive discriminant means that the roots are real, which is confirmed by the expanded discriminant formula \[4(a+b+c)^2 - 12(ab+bc+ac)\].
- Real roots indicate actual intersection points of the quadratic function with the x-axis.
- Distinct (non-repeated) roots occur when the discriminant is greater than zero.
Quadratic Formula
The quadratic formula is a reliable method for finding the solutions to any quadratic equation of the form \(ax^2 + bx + c = 0\). It is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].
This formula uses the discriminant \(b^2 - 4ac\) as its core component, under the square root, to determine if there are real or complex solutions.
This formula uses the discriminant \(b^2 - 4ac\) as its core component, under the square root, to determine if there are real or complex solutions.
- The \(\pm\) symbol indicates the two potential roots: one for addition and one for subtraction, showing the two solutions for \(x\) in case of distinct roots.
- If \(b^2 - 4ac = 0\), both roots converge into a single repeated root due to the square root being zero.
Other exercises in this chapter
Problem 15
The number of solutions (s) of the equation \(\sqrt{3 x^{2}+6 x+7}+\sqrt{5 x^{2}+10 x+14} \leq 4-2 x-x^{2}\) is (A) one (B) two (C) four (D) infinite
View solution Problem 16
If \(\left(a^{2}-1\right) x^{2}+(a-1) x+a^{2}-4 a+3=0\) is an identity in \(x\), then the value of \(a\) is (A) 1 (B) 3 (C) \(-1\) (D) \(-3\)
View solution Problem 19
If \(a, b, c \in R\) and quadratic equation \(x^{2}+(a+b) x+\) \(c=0\) has no real roots then (A) \(c(a+b+c)>0\) (B) \(c+c(a+b+c)>0\) (C) \(c+c(a+b-c)>0\) (D) \
View solution Problem 20
If \(a x^{2}+b x+c=0, a \neq 0, a, b, c \in R\) has distinct real roots in \((1,2)\) then \(a\) and \(5 a+2 b+c\) have (A) same sign (B) opposite sign (C) not d
View solution