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+1 vote
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The below picture 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 load, and so can be found in bridges and in architecture in a variety of forms.

1. 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.

2. If the roots of the quadratic polynomial are equal, where the discriminant D = b– 4ac, then

a) D > 0

b) D < 0

c) D

d) D = 0

3. If α and \(\cfrac1\alpha\) are the zeroes of the quadratic polynomial 2x2 - x + 8k,  then k is

a)  4

b) \(\cfrac14\)

c) \(\cfrac{-1}4\)

d) 2

4. The graph of x+ 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.

5. If the sum of the roots is –p and product of the roots is -\(\cfrac1p\), then the quadratic polynomial is

2 Answers

+1 vote
by (46.5k points)
selected by
 
Best answer

1. c) ‘a’ is a non zero real number and b and c are any real numbers.

For any quadratic polynomial 

ax2 + bx + c

a ≠ 0, and a, b, c are real numbers

2. d) D = 0

If Roots are equal

D = 0

3. b) 1/4

Since α and 1/α are the zeroes of p(x)

Product of Zeroes = c/A

α x 1/α = 8k/2

1 = 4 k

k = 1/4

4. c) Neither touches nor intersects x‐axis.

The graph of x2 + 1 = 0 looks like

Thus, it neither touches nor intersects x-axis.

5. c) k (x2 + px -  1/p)

The quadratic polynomial is

x2 - (Sum of roots)x + Product of roots

Putting values

We can multiply any constant to this polynomial 

So, required quadratic polynomial is

k (x2 + px -  1/p)

+2 votes
by (28.9k points)

1. c) ‘a’ is a non zero real number and b and c are any real numbers.

2. d) D = 0

3. b)

4. c) Neither touches nor intersects x‐axis.

5. c) k (x2 + px - \(\cfrac1p\))

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