Question 11

Rani, a diver was standing on a diving board, 48 feet above the water level. She
took a dive into the pool. Her height above the water level at any time ‘t’ in
seconds is given by the polynomial y(t) such that y(t) = -16 t² + 8t + k.
(i) What is the value of k?
(ii) At what time will she touch the water in the pool?
(iii) Seema’s height above the water level is given by another polynomial h(t) with zeroes -1 and 2. Then, h(t) is given by
(iv) A polynomial q(t) with sum of zeroes as 1 and the product as -6 is modelling Anu’s height in feet above the water at any time t. then, q(t) is given by
(v) The zeroes of the polynomial r(t) = -12 t²+(k – 3) + 48 are negative to each other. Then k is

Solution

(i) Rani’s height at any time is given by y(t) = -16 t² + 8t + k
Initially at t= 0 Rani is standing at a height of 48 feet above the water level. So, at t=0, y = 48.
Substituting all the values k = 48
(ii) When she touches the water level
y(t) = 0
=> -16 t² + 8t + 48 = 0
=> 2 t²– t – 6 = 0
=> (t-2) (2t+ 3) = 0
=> t = 2
(iii) Another polynomial having zeroes as -1 and 2 is given by
h(t) = k (t+1) (t-2)
when time t= 0 height h = 48
so, k (1) (-2) = 48
=> k = – 24
Hence h(t) = -24(t+1)(t-2)
= -24t2 + 24t +48
(iv) the polynomial q(t) will be given by
q(t) = k (t²-t – 6)
at t= 0 height h = 48
so, 48= k (-6)
=> k = – 8
Hence q(t) = -8 (t²-t – 6)
(v) It is given that the zeroes of the polynomial r(t) = -12 t² +(k – 3) + 48 are negative to each other. It means sum of the zeroes will be zero.
So, k-3/ 12 = 0
=> k = 3

Question 12

A highway underpass is parabolic in shape.
(i) it is represented by the polynomial x2 – 2x – 8. Then its zeroes are
(ii) number of the zeroes of the polynomial representing highway underpass is equal to number of points where the graph of polynomial

(iii) Graph of quadratic polynomial is a

(iv) The number of real zeroes that polynomial

Solution

(i) the zeroes are 4 and -2
(ii) intersects x-axis
(iii) parabola
(iv) 0

Question 13

If the zeroes of the polynomial x³– 3x² + x + 1 are in A.P., find the zeroes.

Solution

Let a−b ,a, a+b are zeroes of the polynomial p(x)=x³−3x²+x+1
∴ Sum of its zeroes =(a−b)+a+(a−b) =−(-3) /1⇒3a=3⇒a=1
Sum of the product of its zeroes =a(a−b)+a(a+b)+(a−b)(a+b)=1/1

⇒a²−ab+a²+ab+a²−b²=1

⇒3a²−b²=1⇒3(1)²−b²=1 ⇒b²=3−1=2⇒b=±√2
Since a=1 and b=±√2

So the zeros are 1- √2, 1, 1 +√2 or 1 + √2, 1, 1-√2

Question 14

If a²– 2a – 3 is a factor of a⁴ + pa³ + qa² + 12a – 9 , then find the value of p²– 2q – 3 .

Solution

The value of p²– 2q-3 = 64– 24– 3 = 37

Question 15

If one of the zeroes of the cubic polynomial x³+ ax² + bx + c is -1, then find the product of other two zeroes.

Solution

Zero of f(x) is – 1 so
⇒-1 + a– b + c = 0
⇒ a– b + c = 1
⇒ c = 1 + b–a

⇒βγ = c ⇒βγ = 1 + b–a

Question 16

If the polynomial f(x)=ax³+bx−c is divisible by the polynomial g(x)=x²+bx+c, then find ‘ab’.

Solution

If f(x) is divisible by g(x),then remainder is zero.
x(b-ac+ab²)+(abc-c)for all x
b-ac+ab²=0 and abc-c=0
⇒b-ac+ab²=0 and ab=1

Question 17

If α and β are the zeros of the quadratic polynomial f(x)=x²– p(x + 1)– c,show that (α + 1) (β+1)=1 – c

Solution

Here, sum of the zeroes = 𝛼 + 𝛽 = p
Product of the zeroes =𝛼𝛽 =– (p + c)
LHS = (α + 1) (β + 1) =𝛼𝛽+α + β + 1=– (p + c) + p + 1
=-p-c-p + 1
= 1– c = RHS PROVED

Question 18

If α and β are the zeroes of the polynomial
f(x) = x² + px + q, form a polynomial whose zeroes are (α + β)² and (α – β)²

Solution

The required polynomial is:: ax² +bx +c = x²-2(p² – 2q) x + p²(p²-4q)

Question 19

For what value of k, is the polynomial f(x) = 3x⁴– 9x³ + x² + 15x + k completely divisible
by 3x²– 5?

Solution

Dividing f(x) by g(x),
Given that f(x) is completely divisible by 3x²– 5.
So, the remainder = 0
k + 10 = 0
k = -10

Question 20

If the zeroes of the polynomial x² + px + q are double in value to the zeroes of the polynomial 2x² -5x-3, then find the values of p and q

Solution

Let 𝛼 and𝛽 be the zeros of the polynomial 2x²-5x-3
Then
α + 𝛽 =5/2
And
αβ = – 3 /2
Let 2 α and 2ẞ be the zeros x² + px + q
Then
2α + 2𝛽 = -p
2(α + 𝛽) = -p
2×5/2= -p
p=-5
2α × 2 𝛽 = q
4α 𝛽 = q

SELF ASSESSMENT

Q1. If 𝛼 𝑎𝑛𝑑𝛽 are zeros of quadratic polynomial 2𝑥² + 5𝑥 + 𝑘,then find the value of k such that

(𝛼 + 𝛽)² − 𝛼𝛽 = 24

Q2. If 𝛼 𝑎𝑛𝑑 𝛽 arte the zeroes of 𝑓(𝑥) = 𝑥² + 𝑥 − 2, without finding the zeroes find out the value of

$$(\frac{1}{\alpha }-\frac{1}{\beta })^{2}$$

Q3. Find the zeroes of the quadratic polynomial f(x)= abx² +(b²-ac) x-bc and verify the relationship between zeroes and it’s coefficients.

Q4. If the square of the difference of the zeros of the quadratic polynomial f(x) = x²+ px +45 is equal to 144, find the value of p.

Q5. Divide the polynomial f(x) = 3x² – x³– 3x + 5 by the polynomial g(x) = x – 1 – x² and verify the division algorithm.

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