Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 1.
Construct a quadratic equation with roots 7 and -3.
Solution:
Let the given roots be α = 7 and β = -3
Sum of the roots α + β = 7 + (-3)
α + β = 7 – 3 = 4
Product of the roots αβ = (7)(-3)
αβ = -21
The required quadratic equation is
x2 – (sum of two roots) x + Product of the roots = 0
x2 – 4x – 21 = 0

Question 2.
A quadratic polynomial has one of its zeros 1 + \(\sqrt{5}\) and it satisfies p(1) = 2. Find the quadratic polynomial.
Solution:
Given α = 1 + \(\sqrt{5}\) So, β = 1 – \(\sqrt{5}\)
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 1
The quadratic polynomial is
p(x) = x2 – (α + β)x + αβ
p(x) = k (x2 – 2x – 4)
p( 1) = k(1 – 2 – 4) = -5 k
Given p (1) = 2
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 2

The Polynomial Roots Calculator will find the roots of any polynomial with just one click.

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 3.
If α and β are the roots of the quadratic equation x2 + \(\sqrt{2}\)x + 3 = 0, form a quadratic polynomial with zeroes 1/α, 1/β.
Solution:
α and β are the roots of the equation x2 + \(\sqrt{2}\)x + 3 = 0
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 3
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 4

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 4.
If one root of k(x – 1)2 = 5x – 7 is double the other root, show that k = 2 or – 25.
Solution:
k(x – 1)2 = 5x – 7
(i.e.,) k(x2 – 2x + 1) – 5x + 7 = 0
x2 (k) + x(-2k – 5) + k + 1 = 0
kx2 – x(2k + 5) + (k + 7) = 0
Here it is given that one root is double the other.
So let the roots to α and 2α
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 5
2(4k2 + 25 + 20k) = 9k (k + 7)
2(4k2 + 25 + 20k) = 9k2 + 63k
8k2 + 50 + 40k – 9k2 – 63k = 0
-k2 – 23k + 50 = 0
k2 + 23k – 5o = 0
(k + 25)(k – 2) = 0
k = -25 or 2

Question 5.
If the difference of the roots of the equation 2x2 – (a + 1)x + a – 1 = 0 is equal to their product then prove that a = 2.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 6

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 6.
Find the condition that one of the roots of ax2 + bx + c may be
(i) negative of the other
(ii) thrice the other
(iii) reciprocal of the other.
Solution:
(i) Let the roots be α and -β
Sum of the roots = – b/a = 0 ⇒ b = 0

(ii) Let the roots be α, 3α
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 7

Question 7.
If the equations x2 – ax + b = 0 and x2 – ex + f = 0 have one root in common and if the second equation has equal roots that ae = 2(b + f).
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 8

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 8.
Discuss the nature of roots of
(i) -x2 + 3x + 1 = 0
(ii) 4x2 – x – 2 = 0
(iii) 9x2 + 5x = 0
Solution:
(i) -x2 + 3x + 1 = 0
x2 – 3x – 1 = 0 ———- (1)
Compare this equation with the equation
ax2 + bx + c = 0 ——– (2)
we have a = 1, b = -3, c = -1
Discriminant = b2 – 4ac
b2 – 4ac = (-3)2 – 4 × 1 × – 1
= 9 + 4 =13
b2 – 4ac = 13 > 0
∴ The two roots are real and distinct.

(ii) 4x2 – x – 2 = 0
4x2 – x – 2 = 0 ——(3)
Compare this equation with the equation
ax2 + bx + c = 0 (4)
we have a = 4 , b = – 1, c = – 2
Discriminant = b2 – 4ac
b2 – 4ac = (-1)2 – 4 (4) (-2)
= 1 + 32
= 33
b2 – 4ac = 33 >0
∴ The two roots are real and distinct.

(iii) 9x2 + 5x = 0
9x2 + 5x = 0 ——- (5)
Compare this equation with the equation
ax2 + bx + c = 0 ——– (6)
we have a = 9, b = 5 , c = 0
Discriminant = b2 – 4ac
b2 – 4ac = 52 – 4 × 9 × 0
b2 – 4ac = 25 > 0
∴ The two roots are real and distinct.

Question 9.
Without sketching the graphs find whether the graphs of the following functions will intersect the x- axis and if so in how many points.
(i) y = x2 + x + 2
(ii) y = x2 – 3x – 1
(iii) y = x2 + 6x + 9
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 20
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 21

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 10.
Write f(x) = x2 + 5x + 4 in completed square form.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 22

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 Additional Questions

Question 1.
Find the values of k so that the equation x2 = 2x (1 + 3k) + 7(3 + 2k) = 0 has real and equal roots.
Solution:
The equation is x2 – x(2) (1 + 3k) + 7 (3 + 2k) = 0
The roots are real and equal
⇒ ∆ = 0 (i.e.,) b2 – 4ac = 0
Here a = 1, b = -2 (1 + 3k), c = 7(3 + 2k)
So b2 – 4ac = 0
⇒ [-2 (1 + 3k)]2 – 4(1) (7) (3 + 2k) = 0
(i.e.,) 4 (1 + 3k)2 – 28 (3 + 2k) = 0
(÷ by 4) (1 + 3k)2 – 7(3 + 2k) = 0
1 + 9k2 + 6k – 21 – 14k = 0
9k2 – 8k – 20 = 0
(k – 2)(9k + 10) = 0
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 23
To solve the quadratic inequalities ax2 + bx + c < 0 (or) ax2 + bx + c > 0

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 2.
If the sum and product of the roots of the quadratic equation ax2 – 5x + c = 0 are both equal to 10 then find the values of a and c.
Solution:
The given equation is ax2 – 5x + c = 0
Let the roots be α and β Given α + β = 10 and αβ = 10
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 24

Question 3.
If α and β are the roots of the equation 3x2 – 4x + 1 = 0, form the equation whose roots are \(\frac{\alpha^{2}}{\beta}\) and \(\frac{\beta^{2}}{\alpha}\)
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 25
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 26

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 4.
If one root of the equation 3x2 + kx – 81 = 0 is the square of the other then find k.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 27

Question 5.
If one root of the equation 2x2 – ax + 64 = 0 is twice that of the other then find the value of a.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 28

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