OP Malhotra Class 11 Maths Solutions Chapter 10 Quadratic Equations Exercise 10 (D)

Question 1. Find the condition that one root of \(ax^2 + bx + c = 0\) may be
(i) three times the other,
(ii) it times the other,
(iii) more than the other by h.

Answer:
(i) Let the roots of the given equation \(ax^2 + bx + c = 0\) be \(\alpha\) and \(3\alpha\).
The sum of the roots is given by \(\alpha + 3\alpha = - \frac{b}{a}\).
This simplifies to \(4\alpha = - \frac{b}{a}\).
So, \(\alpha = - \frac{b}{4a}\).
The product of the roots is given by \(\alpha(3\alpha) = \frac{c}{a}\).
This means \(3\alpha^2 = \frac{c}{a}\).
Now, substitute the value of \(\alpha\) into the product of roots equation:
\(3 \left( - \frac{b}{4a} \right)^2 = \frac{c}{a}\)
\(3 \left( \frac{b^2}{16a^2} \right) = \frac{c}{a}\)
\(\frac{3b^2}{16a^2} = \frac{c}{a}\)
Multiply both sides by \(16a^2\) (assuming \(a \ne 0\)):
\(3b^2 = 16ac\).
This is the required condition.
(ii) Let the roots of the given equation \(ax^2 + bx + c = 0\) be \(\alpha\) and \(n\alpha\).
The sum of the roots is \(\alpha + n\alpha = - \frac{b}{a}\).
This gives \((n+1)\alpha = - \frac{b}{a}\).
So, \(\alpha = - \frac{b}{a(n+1)}\).
The product of the roots is \(\alpha(n\alpha) = \frac{c}{a}\).
This means \(n\alpha^2 = \frac{c}{a}\).
Substitute the value of \(\alpha\) into the product of roots equation:
\(n \left( - \frac{b}{a(n+1)} \right)^2 = \frac{c}{a}\)
\(n \left( \frac{b^2}{a^2(n+1)^2} \right) = \frac{c}{a}\)
\(\frac{nb^2}{a^2(n+1)^2} = \frac{c}{a}\)
Multiply both sides by \(a^2(n+1)^2\):
\(nb^2 = ac(n+1)^2\).
This is the required condition.
(iii) Let the roots of the given equation \(ax^2 + bx + c = 0\) be \(\alpha\) and \(\alpha+h\).
The sum of the roots is \(\alpha + (\alpha+h) = - \frac{b}{a}\).
This gives \(2\alpha+h = - \frac{b}{a}\).
\(\implies\) \(2\alpha = - \frac{b}{a} - h\)
\(\implies\) \(\alpha = \frac{1}{2} \left( - \frac{b}{a} - h \right)\).
The product of the roots is \(\alpha(\alpha+h) = \frac{c}{a}\).
Substitute the value of \(\alpha\) into the product of roots equation:
\(\frac{1}{2} \left( - \frac{b}{a} - h \right) \left[ \frac{1}{2} \left( - \frac{b}{a} - h \right) + h \right] = \frac{c}{a}\)
\(\frac{1}{2} \left( \frac{-b-ah}{a} \right) \left[ \frac{-b-ah+2ah}{2a} \right] = \frac{c}{a}\)
\(\frac{1}{2} \left( \frac{-(b+ah)}{a} \right) \left( \frac{ah-b}{2a} \right) = \frac{c}{a}\)
\(\frac{-1}{4a^2} (b+ah)(ah-b) = \frac{c}{a}\)
\(\frac{-1}{4a^2} (a^2h^2-b^2) = \frac{c}{a}\)
\(\frac{b^2-a^2h^2}{4a^2} = \frac{c}{a}\)
Multiply both sides by \(4a^2\):
\(b^2-a^2h^2 = 4ac\)
\(\implies\) \(a^2h^2 = b^2-4ac\).
This is the required condition.
In simple words: For each case, we first write the sum and product of the roots using the given relationships. Then, we substitute the sum-of-roots expression for one root into the product-of-roots expression. This helps us find the special condition that must be true for the coefficients of the quadratic equation.

🎯 Exam Tip: Remember the fundamental relationships between roots and coefficients: for \(Ax^2+Bx+C=0\), sum of roots is \(-B/A\) and product is \(C/A\). These are key to solving such problems.

Question 2. Find the condition that the ratio between the roots of the equation \(ax^2 + bx + c = 0\) may be m: n.
Answer: Let the roots of the equation \(ax^2 + bx + c = 0\) be \(m\alpha\) and \(n\alpha\).
The sum of the roots is \(m\alpha + n\alpha = - \frac{b}{a}\).
\(\implies\) \((m+n)\alpha = - \frac{b}{a}\)
\(\implies\) \(\alpha = - \frac{b}{a(m+n)}\).
The product of the roots is \((m\alpha)(n\alpha) = \frac{c}{a}\).
\(\implies\) \(mn\alpha^2 = \frac{c}{a}\).
Substitute the value of \(\alpha\) into the product of roots equation:
\(mn \left( - \frac{b}{a(m+n)} \right)^2 = \frac{c}{a}\)
\(\implies\) \(mn \left( \frac{b^2}{a^2(m+n)^2} \right) = \frac{c}{a}\)
\(\implies\) \(\frac{mnb^2}{a^2(m+n)^2} = \frac{c}{a}\).
Multiply both sides by \(a^2(m+n)^2\) (assuming \(a \ne 0\) and \(m+n \ne 0\)):
\(\implies\) \(mnb^2 = ac(m+n)^2\).
This is the required condition. This condition relates the coefficients \(a, b, c\) to the ratio \(m:n\) of the roots.
In simple words: If the roots of a quadratic equation are in a specific ratio, we can use the sum and product formulas for roots. By expressing the roots with a common variable and substituting them into the formulas, we can find a condition that links the ratio to the equation's coefficients.

🎯 Exam Tip: Always start by expressing the roots in terms of a common variable when a ratio is given. This simplifies the sum and product of roots calculations.

Question 3. If the ratio of the roots of the equation \(x^2 + px + q = 0\) is equal to the ratio of the roots of \(x^2 + lx + m = 0\), prove that \(mp^2 = ql^2\).
Answer: For the equation \(x^2 + px + q = 0\):
Let the roots be \(\alpha\) and \(\beta\).
Sum of roots: \(\alpha + \beta = -p\)
Product of roots: \(\alpha\beta = q\).
For the equation \(x^2 + lx + m = 0\):
Let the roots be \(\gamma\) and \(\delta\).
Sum of roots: \(\gamma + \delta = -l\)
Product of roots: \(\gamma\delta = m\).
Given that the ratio of the roots is equal:
\(\frac{\alpha}{\beta} = \frac{\gamma}{\delta}\).
Applying the componendo and dividendo rule (which states that if \(\frac{a}{b} = \frac{c}{d}\), then \(\frac{a+b}{a-b} = \frac{c+d}{c-d}\)):
\(\frac{\alpha+\beta}{\alpha-\beta} = \frac{\gamma+\delta}{\gamma-\delta}\).
Square both sides:
\(\frac{(\alpha+\beta)^2}{(\alpha-\beta)^2} = \frac{(\gamma+\delta)^2}{(\gamma-\delta)^2}\).
We know that \((\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta\). Using this identity:
\(\frac{(\alpha+\beta)^2}{(\alpha+\beta)^2 - 4\alpha\beta} = \frac{(\gamma+\delta)^2}{(\gamma+\delta)^2 - 4\gamma\delta}\).
Substitute the sum and product values:
\(\frac{(-p)^2}{(-p)^2 - 4q} = \frac{(-l)^2}{(-l)^2 - 4m}\)
\(\implies\) \(\frac{p^2}{p^2 - 4q} = \frac{l^2}{l^2 - 4m}\).
Cross-multiply:
\(p^2(l^2 - 4m) = l^2(p^2 - 4q)\)
\(\implies\) \(p^2l^2 - 4mp^2 = l^2p^2 - 4ql^2\).
Subtract \(p^2l^2\) from both sides:
\(\implies\) \(-4mp^2 = -4ql^2\).
Divide by -4:
\(\implies\) \(mp^2 = ql^2\).
Hence proved. This result shows a direct relationship between the coefficients of the two quadratic equations when their root ratios are equal.
In simple words: When the way the roots of two different equations relate to each other is the same (like if one root is twice the other in both equations), we can link the numbers in front of 'x' and the constant terms in a special way. This proof uses a trick called componendo and dividendo, which helps combine the sum and product of roots into a single ratio.

🎯 Exam Tip: The identity \((\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta\) is essential when dealing with differences of roots. Componendo and dividendo is a powerful tool for ratios.

Question 4. For what values of a and b, the equation \(x^2 + (2a - 3) x - (3b + 4) = 0\) should have both the roots zero?
Answer: For a quadratic equation in the form \(Ax^2 + Bx + C = 0\) to have both roots equal to zero, two conditions must be met:
1. The sum of the roots must be zero, which means the coefficient of \(x\) (B) must be zero.
2. The product of the roots must be zero, which means the constant term (C) must be zero.
In the given equation, \(x^2 + (2a - 3) x - (3b + 4) = 0\):
The coefficient of \(x\) is \((2a - 3)\). Set this to zero:
\(2a - 3 = 0\)
\(\implies\) \(2a = 3\)
\(\implies\) \(a = \frac{3}{2}\).
The constant term is \( -(3b + 4)\). Set this to zero:
\(-(3b + 4) = 0\)
\(\implies\) \(3b + 4 = 0\)
\(\implies\) \(3b = -4\)
\(\implies\) \(b = - \frac{4}{3}\).
So, for both roots to be zero, \(a\) must be \(\frac{3}{2}\) and \(b\) must be \( - \frac{4}{3}\). This ensures the equation simplifies to \(x^2 = 0\), whose roots are both 0.
In simple words: If both answers (roots) of a 'power of x' equation are zero, it means the part with 'x' and the constant number by itself must both be zero. So, we make those parts equal to zero and solve to find the values of 'a' and 'b'.

🎯 Exam Tip: Remember that for a quadratic equation \(Ax^2+Bx+C=0\), if both roots are zero, then \(B=0\) and \(C=0\). The leading coefficient \(A\) must not be zero.

Question 5. Find the values of \(\lambda\) and \(\mu\) if both the roots of the equation \((3\lambda + 1)x^2 = (2\lambda + 3\mu) x - 3\) are infinite.
Answer: For a quadratic equation \(Ax^2 + Bx + C = 0\) to have infinite roots, the coefficients of the highest power of \(x\) must be zero. Specifically, both the coefficient of \(x^2\) and the coefficient of \(x\) must be zero, while the constant term is non-zero.
First, rewrite the given equation in the standard form \(Ax^2 + Bx + C = 0\):
\((3\lambda + 1)x^2 - (2\lambda + 3\mu)x + 3 = 0\).
Here, \(A = (3\lambda + 1)\), \(B = -(2\lambda + 3\mu)\), and \(C = 3\).
For infinite roots, we set the coefficients \(A\) and \(B\) to zero:
1. Coefficient of \(x^2\) must be zero:
\(3\lambda + 1 = 0\)
\(\implies\) \(3\lambda = -1\)
\(\implies\) \(\lambda = - \frac{1}{3}\).
2. Coefficient of \(x\) must be zero:
\(-(2\lambda + 3\mu) = 0\)
\(\implies\) \(2\lambda + 3\mu = 0\).
Now, substitute the value of \(\lambda = - \frac{1}{3}\) into this equation:
\(2 \left( - \frac{1}{3} \right) + 3\mu = 0\)
\(\implies\) \( - \frac{2}{3} + 3\mu = 0\)
\(\implies\) \(3\mu = \frac{2}{3}\)
\(\implies\) \(\mu = \frac{2}{9}\).
So, the values are \(\lambda = - \frac{1}{3}\) and \(\mu = \frac{2}{9}\). These values make the equation essentially \(3 = 0\), which indicates infinite roots.
In simple words: When the answers (roots) of a quadratic equation are "infinite," it means the parts with \(x^2\) and \(x\) in the equation must both disappear (become zero). We set these parts to zero and solve for the unknown letters \(\lambda\) and \(\mu\).

🎯 Exam Tip: Roots of a quadratic equation become infinite when the coefficients of the higher powers (x² and x) tend to zero, while the constant term remains non-zero. If the constant term also becomes zero, the equation becomes an identity \(0=0\), which means it has infinitely many solutions, not just infinite roots.

Question 6. Find m so that the roots of the equation \(\frac{x^2-b x}{a x-c}=\frac{m-1}{m+1}\) may be equal in magnitude and opposite in sign.
Answer: First, convert the given equation into the standard quadratic form \(Ax^2 + Bx + C = 0\).
The equation is \(\frac{x^2-b x}{a x-c}=\frac{m-1}{m+1}\).
Cross-multiply:
\((x^2 - bx)(m+1) = (m-1)(ax - c)\)
\(\implies\) \((m+1)x^2 - b(m+1)x = a(m-1)x - c(m-1)\)
Rearrange terms to form a quadratic equation:
\((m+1)x^2 + [-b(m+1) - a(m-1)]x + c(m-1) = 0\).
For the roots of a quadratic equation to be equal in magnitude but opposite in sign, their sum must be zero.
The sum of the roots for \(Ax^2 + Bx + C = 0\) is \(-B/A\).
So, we set the sum of the roots to zero:
\(- \frac{[-b(m+1) - a(m-1)]}{(m+1)} = 0\).
This implies that the numerator must be zero:
\(-b(m+1) - a(m-1) = 0\)
\(\implies\) \(-bm - b - am + a = 0\).
Group terms with \(m\):
\(\implies\) \(a - b = am + bm\)
\(\implies\) \(a - b = m(a+b)\).
Finally, solve for \(m\):
\(\implies\) \(m = \frac{a-b}{a+b}\).
This value of \(m\) ensures that the roots are equal in magnitude and opposite in sign. It's important to note that \(a+b\) cannot be zero for \(m\) to be defined in this way.
In simple words: We first change the messy equation into a normal "ax² + bx + c = 0" form. If the roots (answers) are like 5 and -5, their sum is 0. So, we make the middle part (the coefficient of x) of our new equation equal to zero. Then we solve for 'm'.

🎯 Exam Tip: If roots are equal in magnitude and opposite in sign, it implies the sum of roots is zero. Conversely, if the sum of roots is zero and the product of roots is negative, then the roots are real, equal in magnitude, and opposite in sign.

Question 7.
(i) The roots of the quadratic equation \(4x^2 - (5a + 1) x + 5a = 0\), are p and q. If \(q = 1 + p\), calculate the possible values of a, p and q.
(ii) Find the values of p for which the quadratic equation \(x^2 - px + p + 3 = 0\) has
(a) coincident roots,
(b) real distinct roots,
(c) one positive and one negative root.
Answer:
(i) Given the quadratic equation \(4x^2 - (5a + 1) x + 5a = 0\).
The roots are \(p\) and \(q\). We are also given \(q = 1 + p\).
From the properties of quadratic equations:
Sum of roots: \(p+q = - \frac{-(5a+1)}{4} = \frac{5a+1}{4}\).
Substitute \(q = 1+p\):
\(p + (1+p) = \frac{5a+1}{4}\)
\(\implies\) \(2p+1 = \frac{5a+1}{4}\)
\(\implies\) \(2p = \frac{5a+1}{4} - 1 = \frac{5a+1-4}{4} = \frac{5a-3}{4}\)
\(\implies\) \(p = \frac{5a-3}{8}\).
Now find \(q\):
\(q = 1+p = 1 + \frac{5a-3}{8} = \frac{8+5a-3}{8} = \frac{5a+5}{8}\).
Product of roots: \(pq = \frac{5a}{4}\).
Substitute the expressions for \(p\) and \(q\):
\(\left( \frac{5a-3}{8} \right) \left( \frac{5a+5}{8} \right) = \frac{5a}{4}\)
\(\implies\) \(\frac{(5a-3)(5a+5)}{64} = \frac{5a}{4}\)
\(\implies\) \((5a-3)(5a+5) = \frac{64 \cdot 5a}{4}\)
\(\implies\) \(25a^2 + 25a - 15a - 15 = 16 \cdot 5a\)
\(\implies\) \(25a^2 + 10a - 15 = 80a\)
\(\implies\) \(25a^2 - 70a - 15 = 0\).
Divide by 5:
\(\implies\) \(5a^2 - 14a - 3 = 0\).
Factorize the quadratic for \(a\):
\(5a^2 - 15a + a - 3 = 0\)
\(\implies\) \(5a(a-3) + 1(a-3) = 0\)
\(\implies\) \((5a+1)(a-3) = 0\).
So, \(a = 3\) or \(a = - \frac{1}{5}\).
Now, find the corresponding values of \(p\) and \(q\):
Case 1: If \(a=3\)
\(p = \frac{5(3)-3}{8} = \frac{15-3}{8} = \frac{12}{8} = \frac{3}{2}\).
\(q = 1+p = 1 + \frac{3}{2} = \frac{5}{2}\).
Case 2: If \(a = - \frac{1}{5}\)
\(p = \frac{5(- \frac{1}{5})-3}{8} = \frac{-1-3}{8} = \frac{-4}{8} = - \frac{1}{2}\).
\(q = 1+p = 1 - \frac{1}{2} = \frac{1}{2}\).
The possible sets of values are: \((a, p, q) = (3, 3/2, 5/2)\) or \(( -1/5, -1/2, 1/2)\). This demonstrates how a relationship between roots can determine the parameters of the quadratic equation.
(ii) Given the quadratic equation \(x^2 - px + p + 3 = 0\).
Compare with \(Ax^2 + Bx + C = 0\): Here, \(A=1\), \(B=-p\), \(C=p+3\).
The discriminant is \(D = B^2 - 4AC = (-p)^2 - 4(1)(p+3) = p^2 - 4p - 12\).
(a) For coincident roots:
The condition for coincident (equal) roots is that the discriminant \(D\) must be zero.
\(D = 0\)
\(\implies\) \(p^2 - 4p - 12 = 0\).
Factorize the quadratic for \(p\):
\((p-6)(p+2) = 0\).
So, \(p = 6\) or \(p = -2\).
(b) For real distinct roots:
The condition for real and distinct roots is that the discriminant \(D\) must be greater than zero.
\(D > 0\)
\(\implies\) \(p^2 - 4p - 12 > 0\).
From factorization in part (a), \((p-6)(p+2) > 0\).
This inequality holds when \(p < -2\) or \(p > 6\).
(c) For one positive and one negative root:
The condition for roots of opposite signs (one positive, one negative) is that their product must be negative.
The product of roots is \(C/A\).
\(\frac{C}{A} < 0\)
\(\implies\) \(\frac{p+3}{1} < 0\)
\(\implies\) \(p+3 < 0\)
\(\implies\) \(p < -3\).
In simple words: For part (i), we use the sum and product of roots formulas, along with the given relationship between p and q, to form equations. Solving these equations helps us find the possible values for 'a', 'p', and 'q'. For part (ii), we look at the discriminant (D) of the equation. D=0 means equal roots, D>0 means different real roots, and a negative product of roots means one positive and one negative root.

🎯 Exam Tip: Always calculate the discriminant \(D = B^2 - 4AC\) first for questions involving the nature of roots. For a positive and a negative root, ensure the product of roots \(C/A < 0\). Also, be careful with signs when substituting into the discriminant formula.

Question 8. Find the values of m for which the quadratic equation \(x^2 - m (2x - 8) - 15 = 0\) has
(i) equal roots,
(ii) both roots positive.
Answer: First, rewrite the given quadratic equation in the standard form \(ax^2 + bx + c = 0\):
\(x^2 - m(2x - 8) - 15 = 0\)
\(\implies\) \(x^2 - 2mx + 8m - 15 = 0\).
Comparing this with \(ax^2 + bx + c = 0\), we have:
\(a=1\), \(b=-2m\), \(c=8m-15\).
The discriminant is \(D = b^2 - 4ac = (-2m)^2 - 4(1)(8m-15) = 4m^2 - 32m + 60\).
(i) For equal roots:
The condition for equal roots is that the discriminant \(D\) must be zero.
\(D = 0\)
\(\implies\) \(4m^2 - 32m + 60 = 0\).
Divide the entire equation by 4:
\(\implies\) \(m^2 - 8m + 15 = 0\).
Factorize the quadratic for \(m\):
\((m-3)(m-5) = 0\).
So, \(m=3\) or \(m=5\).
(ii) For both roots positive:
For a quadratic equation to have both roots positive, three conditions must be met:
1. The roots must be real, so the discriminant \(D \geq 0\).
2. The sum of the roots must be positive, so \(-b/a > 0\).
3. The product of the roots must be positive, so \(c/a > 0\).
Let's check each condition using the discriminant \(D = 4m^2 - 32m + 60\), which simplifies to \(m^2 - 8m + 15\) when divided by 4.
1. \(D \geq 0\):
From part (i), \(D=0\) when \(m=3\) or \(m=5\). A parabola \(m^2-8m+15\) opens upwards, so \(D \geq 0\) for \(m \leq 3\) or \(m \geq 5\).
2. Sum of roots \(-b/a > 0\):
\(- \frac{-2m}{1} > 0\)
\(\implies\) \(2m > 0\)
\(\implies\) \(m > 0\).
3. Product of roots \(c/a > 0\):
\(\frac{8m-15}{1} > 0\)
\(\implies\) \(8m-15 > 0\)
\(\implies\) \(8m > 15\)
\(\implies\) \(m > \frac{15}{8}\).
Now, we need to find the values of \(m\) that satisfy all three conditions:
- \(m \leq 3\) or \(m \geq 5\)
- \(m > 0\)
- \(m > \frac{15}{8}\) (which is \(m > 1.875\)).
Combining these inequalities: We need \(m\) to be greater than 1.875, and also outside the open interval \((3, 5)\).
So, the possible values for \(m\) are \(\frac{15}{8} < m \leq 3\) or \(m \geq 5\). The constant term \(c\) (which is \(8m-15\)) and \(a\) (which is 1) need to have the same sign, meaning \(c>0\) since \(a>0\).
In simple words: First, we turn the given equation into a standard form. For equal roots, the 'discriminant' (a special calculation from the numbers in the equation) must be zero. For both roots to be positive, the discriminant must be zero or more (for real roots), the sum of roots must be positive, and the product of roots must be positive. We solve for 'm' in each case.

🎯 Exam Tip: When finding conditions for both roots to be positive, remember to check all three conditions: Discriminant \(\geq 0\), Sum of roots \(> 0\), and Product of roots \(> 0\). Missing any one can lead to an incorrect answer.

Question 9. If \(a + b + c = 0\), prove that the roots of \(ax^2 + bx + c = 0\) are rational. Hence, show that the roots of \((p + q) x^2 - 2px + (p - q) = 0\) are rational.
Answer: Part 1: Proving that if \(a+b+c=0\), the roots of \(ax^2 + bx + c = 0\) are rational.
Consider the quadratic equation \(ax^2 + bx + c = 0\).
If we substitute \(x=1\) into the equation:
\(a(1)^2 + b(1) + c = a+b+c\).
Given that \(a+b+c=0\), this means that \(x=1\) is a root of the equation.
Since \(x=1\) is a rational number, and the coefficients \(a, b, c\) are rational (implied for such problems), the other root must also be rational. This is because irrational or complex roots always occur in conjugate pairs; if one root is rational, its pair must also be rational. So, both roots are rational.
Part 2: Showing that the roots of \((p + q) x^2 - 2px + (p - q) = 0\) are rational.
Let's examine the sum of the coefficients for this second quadratic equation:
Coefficient of \(x^2 = (p+q)\)
Coefficient of \(x = (-2p)\)
Constant term \( = (p-q)\)
Sum of coefficients \( = (p+q) + (-2p) + (p-q)\)
\( = p+q-2p+p-q\)
\( = (p-2p+p) + (q-q)\)
\( = 0 + 0 = 0\).
Since the sum of the coefficients is 0, just like in Part 1, \(x=1\) is a root of this equation.
As \(x=1\) is a rational root and the coefficients are rational, the other root must also be rational. Therefore, the roots of \((p + q) x^2 - 2px + (p - q) = 0\) are rational. This property simplifies finding roots significantly.
In simple words: If you add up all the numbers (coefficients) in a quadratic equation and the total is zero, then \(x=1\) is always one of the answers (roots). Since \(x=1\) is a simple number, the other answer must also be a simple number, meaning both roots are 'rational' (can be written as a fraction). We use this rule for both equations.

🎯 Exam Tip: A quick check for \(x=1\) as a root: sum all coefficients. If they add up to zero, then \(x=1\) is definitely a root. This is a common shortcut for showing rationality of roots.

Question 10. Show that the roots of \((x - b) (x - c) + (x - c) (x - a) + (x - a) (x - b) = 0\) are real, and that they cannot be equal unless \(a = b = c\).
Answer: First, expand and simplify the given equation into the standard quadratic form \(Ax^2 + Bx + C = 0\).
\((x - b) (x - c) = x^2 - (b+c)x + bc\)
\((x - c) (x - a) = x^2 - (c+a)x + ca\)
\((x - a) (x - b) = x^2 - (a+b)x + ab\)
Adding these three expressions:
\((x^2 - (b+c)x + bc) + (x^2 - (c+a)x + ca) + (x^2 - (a+b)x + ab) = 0\)
\(\implies\) \(3x^2 - (b+c+c+a+a+b)x + (bc+ca+ab) = 0\)
\(\implies\) \(3x^2 - 2(a+b+c)x + (ab+bc+ca) = 0\).
Now, compare this with \(Ax^2 + Bx + C = 0\):
\(A = 3\)
\(B = -2(a+b+c)\)
\(C = ab+bc+ca\)
To determine the nature of the roots (real or equal), we calculate the discriminant \(D = B^2 - 4AC\).
\(D = [-2(a+b+c)]^2 - 4(3)(ab+bc+ca)\)
\(\implies\) \(D = 4(a+b+c)^2 - 12(ab+bc+ca)\)
Expand \((a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca\):
\(\implies\) \(D = 4(a^2+b^2+c^2+2ab+2bc+2ca) - 12(ab+bc+ca)\)
\(\implies\) \(D = 4a^2+4b^2+4c^2+8ab+8bc+8ca - 12ab-12bc-12ca\)
\(\implies\) \(D = 4a^2+4b^2+4c^2-4ab-4bc-4ca\).
We can factor out 2:
\(\implies\) \(D = 2[2a^2+2b^2+2c^2-2ab-2bc-2ca]\).
Now, rearrange the terms inside the square brackets:
\(\implies\) \(D = 2[(a^2-2ab+b^2) + (b^2-2bc+c^2) + (c^2-2ca+a^2)]\).
Recognize these as perfect squares:
\(\implies\) \(D = 2[(a-b)^2 + (b-c)^2 + (c-a)^2]\).
Since \(a, b, c\) are real numbers, the squares \((a-b)^2\), \((b-c)^2\), and \((c-a)^2\) are always non-negative (greater than or equal to zero).
Therefore, their sum is also non-negative, and \(D = 2 \times (\text{non-negative value}) \geq 0\).
Since the discriminant \(D \geq 0\), the roots of the equation are always real.
For the roots to be equal, the discriminant \(D\) must be exactly zero.
\(D = 0\)
\(\implies\) \(2[(a-b)^2 + (b-c)^2 + (c-a)^2] = 0\).
This implies that the sum of the squares must be zero, which can only happen if each individual square term is zero:
\((a-b)^2 = 0 \implies a-b = 0 \implies a=b\)
\((b-c)^2 = 0 \implies b-c = 0 \implies b=c\)
\((c-a)^2 = 0 \implies c-a = 0 \implies c=a\)
Thus, the roots are equal only when \(a=b=c\). This is a unique and important condition.
In simple words: First, we expand the given equation to get a standard \(Ax^2 + Bx + C = 0\) form. Then we calculate a special number called the 'discriminant'. We found that this number is always positive or zero, which means the answers (roots) are always real numbers. For the answers to be exactly the same, the discriminant must be zero. This only happens if \(a, b,\) and \(c\) are all equal to each other.

🎯 Exam Tip: The identity \(a^2+b^2+c^2-ab-bc-ca = \frac{1}{2}[(a-b)^2+(b-c)^2+(c-a)^2]\) is crucial for proving the reality and equality conditions of roots in such problems. Master this algebraic transformation.

Question 11. Determine the values of m for which the equations \(3x^2 + 4mx + 2 = 0\) and \(2x^2 + 3x - 2 = 0\) may have a common root.
Answer: Let \(\alpha\) be the common root of both quadratic equations.
So, \(\alpha\) must satisfy both equations:
1. \(3\alpha^2 + 4m\alpha + 2 = 0\)
2. \(2\alpha^2 + 3\alpha - 2 = 0\)
First, solve the second equation for \(\alpha\), as it does not contain \(m\):
\(2\alpha^2 + 3\alpha - 2 = 0\)
Factorize the quadratic:
\(2\alpha^2 + 4\alpha - \alpha - 2 = 0\)
\(\implies\) \(2\alpha(\alpha+2) - 1(\alpha+2) = 0\)
\(\implies\) \((2\alpha-1)(\alpha+2) = 0\).
This gives two possible values for \(\alpha\):
\(\alpha = \frac{1}{2}\) or \(\alpha = -2\).
Now, substitute these values of \(\alpha\) into the first equation to find the corresponding values of \(m\).
Case 1: If \(\alpha = \frac{1}{2}\)
Substitute \(\alpha = \frac{1}{2}\) into \(3\alpha^2 + 4m\alpha + 2 = 0\):
\(3 \left( \frac{1}{2} \right)^2 + 4m \left( \frac{1}{2} \right) + 2 = 0\)
\(\implies\) \(3 \left( \frac{1}{4} \right) + 2m + 2 = 0\)
\(\implies\) \(\frac{3}{4} + 2m + 2 = 0\)
\(\implies\) \(2m = -2 - \frac{3}{4}\)
\(\implies\) \(2m = - \frac{8}{4} - \frac{3}{4} = - \frac{11}{4}\)
\(\implies\) \(m = - \frac{11}{8}\).
Case 2: If \(\alpha = -2\)
Substitute \(\alpha = -2\) into \(3\alpha^2 + 4m\alpha + 2 = 0\):
\(3(-2)^2 + 4m(-2) + 2 = 0\)
\(\implies\) \(3(4) - 8m + 2 = 0\)
\(\implies\) \(12 - 8m + 2 = 0\)
\(\implies\) \(14 - 8m = 0\)
\(\implies\) \(8m = 14\)
\(\implies\) \(m = \frac{14}{8} = \frac{7}{4}\).
Thus, the values of \(m\) for which the equations have a common root are \( - \frac{11}{8}\) and \(\frac{7}{4}\). A common root implies that there is a value of x that satisfies both equations simultaneously.
In simple words: To find 'm' when two equations share a root, first solve the equation that doesn't have 'm' in it to find its roots. Then, take each of those roots and plug it into the other equation that contains 'm'. This will give you the possible values for 'm'.

🎯 Exam Tip: When finding common roots, solve the equation with known coefficients first. Then, substitute those roots into the second equation to find the unknown parameter. This method is often simpler than using cross-multiplication for elimination.

Question 12. Find the value of k, so that the equation \(2x^2 + kx - 5 = 0\) and \(x^2 - 3x - 4 = 0\) may have one root common.
Answer: Let \(\alpha\) be the common root of both quadratic equations.
So, \(\alpha\) must satisfy both equations:
1. \(2\alpha^2 + k\alpha - 5 = 0\)
2. \(\alpha^2 - 3\alpha - 4 = 0\)
First, solve the second equation for \(\alpha\):
\(\alpha^2 - 3\alpha - 4 = 0\)
Factorize the quadratic:
\((\alpha-4)(\alpha+1) = 0\).
This gives two possible values for \(\alpha\):
\(\alpha = 4\) or \(\alpha = -1\).
Now, substitute these values of \(\alpha\) into the first equation to find the corresponding values of \(k\).
Case 1: If \(\alpha = 4\)
Substitute \(\alpha = 4\) into \(2\alpha^2 + k\alpha - 5 = 0\):
\(2(4)^2 + k(4) - 5 = 0\)
\(\implies\) \(2(16) + 4k - 5 = 0\)
\(\implies\) \(32 + 4k - 5 = 0\)
\(\implies\) \(27 + 4k = 0\)
\(\implies\) \(4k = -27\)
\(\implies\) \(k = - \frac{27}{4}\).
Case 2: If \(\alpha = -1\)
Substitute \(\alpha = -1\) into \(2\alpha^2 + k\alpha - 5 = 0\):
\(2(-1)^2 + k(-1) - 5 = 0\)
\(\implies\) \(2(1) - k - 5 = 0\)
\(\implies\) \(2 - k - 5 = 0\)
\(\implies\) \(-3 - k = 0\)
\(\implies\) \(k = -3\).
Thus, the values of \(k\) for which the equations have a common root are \( - \frac{27}{4}\) and \(-3\). Having a common root means there's an \(x\) value that works for both equations.
In simple words: We find the answers (roots) for the simpler quadratic equation first. Then, we take each of these answers and put it into the other equation (the one with 'k'). This helps us find the possible values for 'k' that make both equations true for that common answer.

🎯 Exam Tip: When one equation has all numerical coefficients, solve it directly to find the possible common roots. This simplifies the substitution process significantly compared to using cross-multiplication if not comfortable with it.

Question 13. If \(ax^2 + bx + c = 0\) and \(bx^2 + cx + a = 0\) have a common root, prove that \(a + b + c = 0\) or \(a = b = c\).
Answer: Let \(\alpha\) be the common root of the two quadratic equations:
1. \(a\alpha^2 + b\alpha + c = 0\)
2. \(b\alpha^2 + c\alpha + a = 0\)
We can solve for \(\alpha^2\), \(\alpha\), and 1 using the cross-multiplication method for systems of linear equations (treating \(\alpha^2\) and \(\alpha\) as variables):
\(\frac{\alpha^2}{b \cdot a - c \cdot c} = \frac{\alpha}{c \cdot b - a \cdot a} = \frac{1}{a \cdot c - b \cdot b}\)
\(\implies\) \(\frac{\alpha^2}{ab - c^2} = \frac{\alpha}{bc - a^2} = \frac{1}{ac - b^2}\).
From the first two parts, we can find \(\alpha\):
\(\alpha = \frac{ab - c^2}{bc - a^2}\).
From the last two parts, we can also find \(\alpha\):
\(\alpha = \frac{bc - a^2}{ac - b^2}\).
Equating the two expressions for \(\alpha\):
\(\frac{ab - c^2}{bc - a^2} = \frac{bc - a^2}{ac - b^2}\).
Cross-multiply:
\((ab - c^2)(ac - b^2) = (bc - a^2)^2\).
Expand both sides:
\(a^2bc - ab^3 - ac^3 + b^2c^2 = b^2c^2 - 2a^2bc + a^4\).
Subtract \(b^2c^2\) from both sides:
\(a^2bc - ab^3 - ac^3 = -2a^2bc + a^4\).
Rearrange terms to one side:
\(a^4 + ab^3 + ac^3 - 3a^2bc = 0\).
Factor out \(a\) (assuming \(a \neq 0\); if \(a=0\), the original equation isn't quadratic):
\(a(a^3 + b^3 + c^3 - 3abc) = 0\).
Since \(a \neq 0\), we must have:
\(a^3 + b^3 + c^3 - 3abc = 0\).
Recall the algebraic identity: \(a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)\).
So, \((a+b+c)(a^2+b^2+c^2-ab-bc-ca) = 0\).
This implies that either of the factors must be zero:
Case 1: \(a+b+c = 0\).
Case 2: \(a^2+b^2+c^2-ab-bc-ca = 0\).
The second factor can be rewritten using another identity: \(a^2+b^2+c^2-ab-bc-ca = \frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2]\).
So, \(\frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2] = 0\).
This equation holds if and only if each squared term is zero (since squares of real numbers are non-negative):
\((a-b)^2 = 0 \implies a=b\)
\((b-c)^2 = 0 \implies b=c\)
\((c-a)^2 = 0 \implies c=a\)
Thus, Case 2 leads to \(a=b=c\).
Therefore, if the two equations have a common root, then either \(a+b+c=0\) or \(a=b=c\). This is a classical result that links shared roots to the coefficients.
In simple words: We assume both equations share an answer (root). Using a method called cross-multiplication, we find two different ways to write this common answer using 'a', 'b', and 'c'. Setting these two expressions equal leads us to a well-known math rule. This rule tells us that either the sum of 'a', 'b', 'c' is zero, or all three numbers 'a', 'b', 'c' must be the same.

🎯 Exam Tip: Problems involving a common root between two general quadratic equations often require the cross-multiplication method for elimination. Remember the algebraic identity for \(a^3+b^3+c^3-3abc\) to simplify the derived condition. Be careful with calculations during expansion.

Question 14. The equations \(x^2 + x + a = 0\) and \(x^2 + ax + 1 = 0\) have a common real root
(a) for no value of a
(b) for exactly one value of a
(c) for exactly two values of a
(d) for exactly three values of a
Answer: (b) for exactly one value of a
Let \(\alpha\) be the common root of the two equations:
1. \(\alpha^2 + \alpha + a = 0\)
2. \(\alpha^2 + a\alpha + 1 = 0\)
Subtract equation (2) from equation (1):
\((\alpha^2 + \alpha + a) - (\alpha^2 + a\alpha + 1) = 0\)
\(\implies\) \(\alpha - a\alpha + a - 1 = 0\)
\(\implies\) \(\alpha(1-a) + (a-1) = 0\)
\(\implies\) \(\alpha(1-a) - (1-a) = 0\)
\(\implies\) \((1-a)(\alpha - 1) = 0\).
This implies two possibilities:
Case 1: \(1-a = 0 \implies a = 1\).
If \(a=1\), both original equations become \(x^2+x+1=0\).
The discriminant for this equation is \(D = 1^2 - 4(1)(1) = 1-4 = -3\).
Since \(D < 0\), the roots are complex (non-real). Therefore, \(a=1\) does not lead to a common *real* root.
Case 2: \(\alpha - 1 = 0 \implies \alpha = 1\).
If \(\alpha = 1\) is the common root, substitute \(x=1\) into either of the original equations to find \(a\). Using the first equation:
\((1)^2 + (1) + a = 0\)
\(\implies\) \(1 + 1 + a = 0\)
\(\implies\) \(2 + a = 0\)
\(\implies\) \(a = -2\).
Now, we must verify that for \(a=-2\), the roots are indeed real.
The first equation becomes \(x^2+x-2=0\). Its roots are \((x+2)(x-1)=0\), which gives \(x=1, -2\) (both real).
The second equation becomes \(x^2-2x+1=0\). Its roots are \((x-1)^2=0\), which gives \(x=1, 1\) (both real).
Both equations have real roots and share \(x=1\) as a common real root when \(a=-2\).
Therefore, there is exactly one value of \(a\) (which is \(a=-2\)) for which the equations have a common real root.
In simple words: We assume there is a common answer (root) for both equations. By subtracting one equation from the other, we find that either 'a' must be 1, or the common root must be 1. If 'a' is 1, the equations give complex numbers, not real ones. If the common root is 1, we find 'a' to be -2. When 'a' is -2, both equations have real answers. So, there is only one value for 'a'.

🎯 Exam Tip: When equations share a common root, subtracting them often simplifies to a linear equation that helps find either the root or the unknown parameter. Always remember to check the condition for "real" roots (Discriminant \(\geq 0\)) for the parameter values found.