## Class 10 Chapter 2(Poloynomials)

### Case based 4:

**The pictures are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures,their curve represents an efficient method of carrying load and so can be found in bridges and in architecture in a variety of forms.(Class 10 Case based problem of Chapter 2 Polynomials 3)**

**(A) In the standard form of quadratic polynomial, ax² + bx + c, a, b and c are :**

**(a) All are real numbers.**

**(b) All are rational numbers.**

**(c) ‘a’ is a non-zero real number and, b and c are any real numbers.**

**(d) All are integers.**

**(B) If the root of quadratic polynomialare equal, where the discrimnant D = b² – 4ac, then:**

**(a) D > 0 (b) D < 0**

**(c) D ≥ 0 (d) D = 0**

**(C) If α and 1/α are the zeroes of the quadratic polynomial 2x² – x + 8k, then k is:**

**(a) 4 (b) 1/4**

**(c) -1/4 (d) 2**

**(D) The graph of x² + 1 = 0:**

**(a) Intersects x-axis at two distinct points**

**(b) Touches x – axis at a point.**

**(c) Neither touches nor intersects x – axis**

**(d) Either touches or intersects x- axis.**

**(E) If the sum of the roots is -p and product of the roots is -1/p, then then the quadratic polynomial is**

**(a) k(-px² + x/p + 1)**

**(b) k(px² – x/p – 1)**

**(c) k(x² + px – 1/p)**

**(d) k(x² – px + 1/p)**

## Solution:

**(A) Answer (c) ‘a’ is non-zero real number and b and c are any real numbers.**

**Explatnation:** The standard form of the quadratic polynomial is **ax² + bx + c,** where a, b, c are real numbers and a ≠ 0

**(B) Answer (d) D = 0**

**(C) Answer(b) k = 1/4**

**Explanation:** **2x² – x + 8k**

Roots of polynomial = α, 1/α

Product of the roots = Constant term/ Coefficient of x²

⇒ α× 1/α = 8k/2

⇒ k = 1/4

(D) Answer (c) Neither touches nor intersects x – axis,

**Explanation:** **x² + 1 = 0**

⇒ x² = -1

It has no real roots

Hence the graph of equation never touches and intersect the x- axis

**(E) Answer (c**) **k(x² + px – 1/p)**

**Explanation: **sum of the roots = -p

product of the roots = -1/p

Quadratic equation

= k[x² – (Sum of the roots) x + (Productof the roots)]

= k[x² – (-p)x +(-1/p) ]

= k[x² + px -1/p]

# Case Based 5:

Government of India allotted some funds to the National Disaster Relief Fund to help the families of flood affected village. The fund alloted is represented by . The fund is equally divided between each of the families of that village. Each family received an amount of . After distribution, the amount left was . Which was used to construct a primary health centre in each village.

**(A)** If the product of the zeros of the polynomial is 4, then then the value of a is:

(a) (b)

(c) -1 (d) 0

**(B)** If α and β are zeroes of the quadratic polynomial such that α² + β² + αβ = 21/4, The value of k is

(a) -2 (b) 0

(c) (d)

**(C)** If the sum of squares of zeroes of the polynomial is 40, the value of k is

(a) 2 (b) 4

(c) 8 (d) 12

**(D)** The zeroes of the quadratic polynomial are:

(a) -a, -1/a (b) a, 1/a

(c) -1, a (d) 1, -a

**(E)** If one of the zeroes of the quadratic polynomial is 2, then the value of a is:

(a) 10 (b) -10

(c) -7 (d) -2

### Solution:

**(A) Answer (a)**

Explanation: Given

Product of the zeroes =

**(B) Answer (c) -5/2**

Explanation:Given

Since,

Given that, α² + β² + αβ = 21/4

add an substract αβ

α² + β² + αβ + αβ- αβ = 21/4

⇒α² + β² + 2αβ – αβ = 21/4

⇒(α + β)²-αβ = 21/4

**(C) Answer (d) 12**

Explanation: Let zeroes of the equation are α and β

Then, α + β = -b/a = -(8)/1 =8

α.β = c/a = k/1 = k

Given that

**(D) Answer (b) a, 1/a**

Explanation: given that

Hence, zeroes are = a, 1/a

**(E)Answer (b) -10**

Let zeroes of the equation are α and β

Given

and