Quadratic Equations
Section A: Multiple Choice Questions
1. If the quadratic equation \(3x + \frac{1}{x} = x + 3\) is written in standard form, then:
Solution Multiply by x: \(3x^2 + 1 = x^2 + 3x\). Rearrange: \(2x^2 - 3x + 1 = 0\). Comparing with \(ax^2+bx+c=0\), we get \(a=2, b=-3, c=1\).
2. For the equation \(x^2 + 5x - 1\), which of the following statements is correct?
Solution \(D = b^2 - 4ac = 5^2 - 4(1)(-1) = 25 + 4 = 29\). Since \(D > 0\) and 29 is not a perfect square, roots are real, distinct, and irrational.
3. If the roots of the equation \(ax^2 + bx + c\) are real and equal, what will be the relation between a, b, c?
Solution For equal roots, Discriminant \(D = 0 \Rightarrow b^2 - 4ac = 0 \Rightarrow b^2 = 4ac\).
4. If \(\frac{1}{2}\) is a root of the quadratic equation \(x^2 - mx - \frac{5}{4} = 0\), then value of m is:
Solution Substitute \(x = 1/2\): \((\frac{1}{2})^2 - m(\frac{1}{2}) - \frac{5}{4} = 0 \Rightarrow \frac{1}{4} - \frac{m}{2} - \frac{5}{4} = 0 \Rightarrow -1 - \frac{m}{2} = 0 \Rightarrow m = -2\).
5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, the other two sides are:
Solution Let base = x. Altitude = x-7. \(x^2 + (x-7)^2 = 13^2 \Rightarrow 2x^2 - 14x + 49 = 169 \Rightarrow 2x^2 - 14x - 120 = 0 \Rightarrow x^2 - 7x - 60 = 0\). Factors are -12, +5. So x = 12. Altitude = 5.
6. The sum of a number and its reciprocal is \(\frac{65}{8}\). Then the number is:
Solution \(x + \frac{1}{x} = \frac{65}{8} \Rightarrow \frac{x^2+1}{x} = \frac{65}{8} \Rightarrow 8x^2 - 65x + 8 = 0\). Factors are -64, -1. \(x(8x-1) - 8(8x-1) = 0\). So x = 8 or 1/8.
7. If one root of equation \(4x^2 - 2x + k - 4 = 0\) is reciprocal of the other, the value of k is:
Solution Product of roots \(\alpha \cdot \frac{1}{\alpha} = 1\). Product formula = \(c/a\).
\(\frac{k-4}{4} = 1 \Rightarrow k - 4 = 4 \Rightarrow k = 8\).
\(\frac{k-4}{4} = 1 \Rightarrow k - 4 = 4 \Rightarrow k = 8\).
8. The value of \(\sqrt{6 + \sqrt{6 + \sqrt{6 \dots}}}\) is
Solution Let \(x = \sqrt{6+x} \Rightarrow x^2 = 6+x \Rightarrow x^2 - x - 6 = 0\).
\((x-3)(x+2) = 0\). Since value must be positive, x = 3.
\((x-3)(x+2) = 0\). Since value must be positive, x = 3.
9. Which of the following is not a quadratic equation?
Solution A quadratic equation must have degree 2. Option (b) has degree 3, so it is a cubic equation.
10. The equation \(2x^2 + kx + 3 = 0\) has two equal roots, then the value of k is
Solution For equal roots, \(b^2 - 4ac = 0 \Rightarrow k^2 - 4(2)(3) = 0 \Rightarrow k^2 = 24\).
\(k = \pm \sqrt{24} = \pm 2\sqrt{6}\).
\(k = \pm \sqrt{24} = \pm 2\sqrt{6}\).
Section B: Assertion-Reason Questions
Directions:
(a) Both A and R are true and R is correct explanation of A.
(b) Both A and R are true but R is not correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
(a) Both A and R are true and R is correct explanation of A.
(b) Both A and R are true but R is not correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
11. Assertion (A): \(3x^2 - 6x + 3 = 0\) has repeated roots.
Reason (R): The quadratic equation \(ax^2 + bx + c = 0\) have repeated roots if discriminant D > 0.
Reason (R): The quadratic equation \(ax^2 + bx + c = 0\) have repeated roots if discriminant D > 0.
Solution For Assertion: \(D = (-6)^2 - 4(3)(3) = 36 - 36 = 0\). Since D=0, roots are repeated (True).
For Reason: Repeated roots require D = 0, not D > 0 (False).
For Reason: Repeated roots require D = 0, not D > 0 (False).
12. Assertion (A): If one root of \(6x^2 - x - k = 0\) is \(2/3\), then the value of k is 2.
Reason (R): The quadratic equation \(ax^2 + bx + c = 0, a \neq 0\) has almost two roots.
Reason (R): The quadratic equation \(ax^2 + bx + c = 0, a \neq 0\) has almost two roots.
Solution Substitute \(x = 2/3\): \(6(4/9) - 2/3 - k = 0 \Rightarrow 8/3 - 2/3 = k \Rightarrow k = 6/3 = 2\) (True). Reason is a correct fact but doesn't explain the value of k.
13. Assertion (A): \((2x - 1)^2 - 4x^2 + 5 = 0\) is not a quadratic equation.
Reason (R): An equation of the form \(ax^2 + bx + c = 0, a \neq 0\) is called a quadratic equation.
Reason (R): An equation of the form \(ax^2 + bx + c = 0, a \neq 0\) is called a quadratic equation.
Solution Expanding Assertion: \(4x^2 - 4x + 1 - 4x^2 + 5 = 0 \Rightarrow -4x + 6 = 0\). This is linear, not quadratic (True). Reason correctly defines quadratic (True) and explains why the Assertion is true (because 'a' becomes 0).
14. Assertion (A): The roots of the quadratic equation \(x^2 + 2x + 2 = 0\) are imaginary.
Reason (R): If discriminant \(D = b^2 - 4ac < 0\) then the roots are imaginary.
Reason (R): If discriminant \(D = b^2 - 4ac < 0\) then the roots are imaginary.
Solution \(D = 2^2 - 4(1)(2) = 4 - 8 = -4 < 0\). Since D is negative, roots are imaginary. Reason is correct explanation.
15. Assertion (A): The quadratic equation \(x^2 + 4x + 5 = 0\) has no real roots.
Reason (R): The discriminant is greater than zero.
Reason (R): The discriminant is greater than zero.
Solution \(D = 16 - 20 = -4 < 0\). So no real roots (True). Reason says D > 0, which is False.
16. Assertion (A): The solution of the quadratic equation \((x-1)^2 - 5(x-1) - 6 = 0\) is 7.
Reason (R): The solution of equation \(x^2 + 5x - (a^2+a-6) = 0\) is a+3.
Reason (R): The solution of equation \(x^2 + 5x - (a^2+a-6) = 0\) is a+3.
Solution Let \(y=x-1\). \(y^2-5y-6=0 \Rightarrow (y-6)(y+1)=0 \Rightarrow y=6\). \(x-1=6 \Rightarrow x=7\). Assertion is True. Reason is an unrelated calculation.
17. Assertion (A): Every quadratic equation has exactly one root.
Reason (R): Every quadratic equation has at most two roots.
Reason (R): Every quadratic equation has at most two roots.
Solution Quadratic equations have 2 roots (real or imaginary). Assertion is False. Reason is True.
18. Assertion (A): If the coefficient of \(x^2\) and the constant term have opposite signs, the quadratic equation has real roots.
Reason (R): Every quadratic equation has at least two roots.
Reason (R): Every quadratic equation has at least two roots.
Solution If 'a' and 'c' have opposite signs, \(ac < 0\). Then \(D = b^2 - 4ac\) becomes \(b^2 + \text{positive}\), which is always positive. So real roots (True). Reason is False (can have zero real roots).
19. Assertion (A): The equation \(x^2 + 4x + 5 = 0\) has two equal roots.
Reason (R): If discriminant \(D = b^2 - 4ac = 0\), then roots are equal.
Reason (R): If discriminant \(D = b^2 - 4ac = 0\), then roots are equal.
Solution \(D = 16 - 20 = -4 \neq 0\). Roots are not equal. Assertion is False. Reason is True statement.
20. Assertion (A): The roots of the quadratic equation \(x^2 + 2x + 2 = 0\) are imaginary.
Reason (R): If discriminant \(D = b^2 - 4ac < 0\) then roots are imaginary.
Reason (R): If discriminant \(D = b^2 - 4ac < 0\) then roots are imaginary.
Solution \(D = 4 - 8 = -4 < 0\). Assertion is True. Reason correctly explains it.