Q) 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, a x² + b x + 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 = b² – 4ac, then
a) D > 0                    b) D < 0                      c) D >= 0                        d) D = 0

3. If and are the zeroes of the qudratic polynomial 2 x² – x + 8 k, then k is
a) 4                    b)                      c)                          d) 2

4. 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

5. If the sum of the roots is – p and product of the roots is , then the quadratic polynomial is
a) k( – p X² + + 1)            b)  k( p X² – – 1)
c) k( X² + p X – )                         d) k( X² + p X + ) 

Ans:

STEP BY STEP SOLUTION

1. Nature of a, b, c

We know that a standard quadratic polynomial expression is: a x² + b x + c = 0

here, a ≠ 0, and a, b, c are real numbers.

Therefore, option c) is correct

2. Nature of D:  

We know that in a standard quadratic polynomial expression, a x² + b x + c = 0, the discriminant is expressed by D = b² – 4ac.

In this polynomial equation, when roots are equal, Discriminant is zero.

∴ b² – 4ac = 0 or D = 0

Therefore, option d) is correct

3. Value of k:

The given Quadratic polynomial equation: 2 x² – x + 8 k

When we compare with the standard quadratic polynomial expression, a x² + b x + c, we get:

a = 2, b = – 1, c = 8k

We know that in a standard quadratic polynomial:

Sum of roots, …. (i)

and Product of the roots, …. (ii)

By substituting the values of c and a in equation (ii),

∴ 1 = 4 k

∴ k =

Therefore, option b) is correct

4. Nature of graph:

Let’s start by finding values of the graph for different values of x and plot it

To plot it, we consider equation of graph as f(x) = x² + 1 and plot values of f(x) on y axis.

for x = 0,  f(x) = (0)² + 1 = 1 

for x = 1, f(x) = (1)² + 1 = 2

for x = 2,  f(x) = (2)² + 1 = 5

for x = – 1, f(x) = (- 1)² + 1 = 2

for x = – 2, f(x) = (- 2)² + 1 = 5

Since on x-axis, f(x) = 0, and for no value of x, its value of f(x) is 0, hence the graph doesn’t touch X – axis, nor intersects X-axis.

Therefore, option c) is correct

5. Equation for Quadratic Polynomial: 

We know that a standard Quadratic Polynomial equation is expressed as:

X²  – (sum of roots) X + (products of roots) = 0

Given that sum of roots = – p

and product of roots = –

Therefore, the equation will be: X² – (- p) X + (- )

= X² + p X –

And if we multiply, it by any constant k, we get:

k (X² + p X – )

Therefore, option c) is correct

Please press the “Heart” button if you like the solution.