CBSE Class 9 Mathematics Quadratic Equations MCQs Set 09

Answer: (a) \( \left\{1+\sqrt{39.5},\ 1-\sqrt{39.5}\right\} \)

Explanation:
1. Write \( 2x^2-4x-77=0 \): \( x=\dfrac{4\pm\sqrt{16+616}}4=\dfrac{4\pm\sqrt{632}}4 \).
2. \( \sqrt{632}=2\sqrt{158} \), so \( x=1\pm\dfrac{\sqrt{158}}2=1\pm\sqrt{39.5} \).

Teacher's Note:
1. Use the formula with a = 2, b = -4, c = -77.
2. The discriminant is 16 + 616 = 632.

Answer: (c) \( -\dfrac12\text{ and }\dfrac23 \)

Explanation:
1. \( 6x^2-4x+3x-2=0 \), i.e. \( 2x(3x-2)+1(3x-2)=0 \).
2. \( (2x+1)(3x-2)=0 \), so \( x=-\dfrac12 \) or \( x=\dfrac23 \).

Teacher's Note:
1. Split -x as -4x + 3x (product -12).
2. Check: sum of roots 1/6 = -(-1)/6.

Answer: (d) \( \text{one root, }x=\dfrac12 \)

Explanation:
1. Multiplying by 12: \( 6(x^2-3x+2)-(2x^2-1)=4x(x+2) \), i.e. \( 4x^2-18x+13=4x^2+8x \).
2. The \( x^2 \) terms cancel: \( 26x=13 \), so the only root is \( x=\dfrac12 \).

Teacher's Note:
1. Expand carefully; the quadratic terms may cancel.
2. This reduces to a linear equation.

Answer: (c) 24

Explanation:
1. Length 600 mm in \( n \) parts: \( \left(\dfrac{600}n-1\right)(n+1)=600 \).
2. \( n^2+n-600=0 \), so \( (n-24)(n+25)=0 \); reject \( n=-25 \), so \( n=24 \).

Teacher's Note:
1. Number of parts times length of one part is constant (600 mm).
2. 25 parts of 24 mm = 600 mm.

Answer: (d) 18 km/h and 20 km/h

Explanation:
1. Shorter-route speed \( x \): \( \dfrac{81}x-\dfrac{85}{x+2}=\dfrac14 \), which simplifies to \( x^2+18x-648=0 \)
2. \( (x-18)(x+36)=0 \), so \( x=18 \) (reject \( -36 \)); the other car goes at 20 km/h.

Teacher's Note:
1. 15 minutes is one quarter of an hour.
2. Times: 4.5 h and 4.25 h.

Answer: (c) 21 and 23 (or -23 and -21)

Explanation:
1. Let them be \( 2n-1 \) and \( 2n+1 \): \( 2(4n^2+1)=970 \).
2. \( n^2=121 \), so \( n=\pm11 \) and the numbers are \( 21,23 \) or \( -23,-21 \).

Teacher's Note:
1. Consecutive odd numbers differ by 2.
2. 441 + 529 = 970.

Answer: (d) 49

Explanation:
1. Dull days \( n \): \( n+(n+4)^2+4=56 \), i.e. \( n^2+9n-36=0 \).
2. \( (n+12)(n-3)=0 \), so \( n=3 \) and wet days \( =(3+4)^2=49 \).

Teacher's Note:
1. Add the three kinds of days to get 56.
2. 3 + 49 + 4 = 56.

Answer: (b) not real (imaginary)

Explanation:
1. \( \Delta=2^2-4(1)(4)=-12\lt 0 \).
2. A negative discriminant means the roots are not real.

Teacher's Note:
1. A square root of a negative number is not real.
2. The parabola y = x^2 + 2x + 4 never meets the x-axis.

Answer: (d) equal, each equal to \( -\dfrac32 \)

Explanation:
1. \( \Delta=144-144=0 \), so the roots are equal.
2. \( 4x^2+12x+9=(2x+3)^2=0 \) gives \( x=-\dfrac32 \) twice.

Teacher's Note:
1. Zero discriminant gives a repeated root of -b/2a.
2. -12/8 = -3/2.

Answer: (b) \( -4;\ 0\text{ and }4 \)

Explanation:
1. \( y=(x-2)^2-4 \), so the minimum value is \( -4 \) at \( x=2 \).
2. The curve cuts the x-axis where \( x(x-4)=0 \), i.e. at \( x=0 \) and \( x=4 \).

Teacher's Note:
1. The graph is a parabola opening upward through the origin.
2. Table: y = 12, 5, 0, -3, -4, -3, 0, 5 for x = -2 to 5.

Answer: (c) \( -4\text{ and }3 \)

Explanation:
1. Put \( y=0 \): \( x^2+x-12=(x+4)(x-3)=0 \).
2. So the roots are \( x=-4 \) and \( x=3 \).

Teacher's Note:
1. The roots of the equation are the x-intercepts of the curve.
2. Table values: y = 0 at x = -4 and x = 3.

Answer: (b) \( 11 \)

Explanation:
1. \( 7+4x-x^2=11-(x-2)^2 \).
2. The maximum value is \( 11 \), reached at \( x=2 \).

Teacher's Note:
1. Complete the square: -(x - 2)^2 + 11.
2. Table: y = 7, 10, 11, 10, 7 for x = 0, 1, 2, 3, 4.

Answer: (c) \( -9;\ 2\text{ and }-4 \)

Explanation:
1. \( y=(x+1)^2-9 \), so the minimum is \( -9 \) at \( x=-1 \).
2. \( x^2+2x-8=(x+4)(x-2)=0 \) gives the roots \( -4 \) and \( 2 \).

Teacher's Note:
1. Complete the square to find the least value.
2. Table: y = 7, -5, -9, -8, -5, 7 for x = -5, -3, -1, 0, 1, 3.

Answer: (b) \( a=\dfrac12,\ c=5 \)

Explanation:
1. Sum \( =\dfrac5a=10 \), so \( a=\dfrac12 \).
2. Product \( =\dfrac ca=10 \), so \( c=10a=5 \).

Teacher's Note:
1. Apply sum = -b/a and product = c/a.
2. Check: 5/(1/2) = 10 and 5/(1/2) = 10.

Answer: (d) \( k=6 \) or \( k=-6 \)

Explanation:
1. Equal roots need \( \Delta=k^2-36=0 \).
2. So \( k=\pm6 \); the roots are \( 3,3 \) or \( -3,-3 \).

Teacher's Note:
1. Set the discriminant equal to zero.
2. Check: x^2 - 6x + 9 = (x - 3)^2.

Answer: (c) \( x^2-x-6=0 \)

Explanation:
1. Sum \( =3+(-2)=1 \) and product \( =3\times(-2)=-6 \).
2. The equation is \( x^2-x-6=0 \).

Teacher's Note:
1. Use x^2 - (sum)x + product.
2. Check: 9 - 3 - 6 = 0 and 4 + 2 - 6 = 0.