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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

x^{2} – (sum of two roots) x + Product of the roots = 0

x^{2} – 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 – \(\sqrt{5}\)

The quadratic polynomial is

p(x) = x^{2} – (α + β)x + αβ

p(x) = k (x^{2} – 2x – 4)

p( 1) = k(1 – 2 – 4) = -5 k

Given p (1) = 2

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

Question 3.

If α and β are the roots of the quadratic equation x^{2} + \(\sqrt{2}\)x + 3 = 0, form a quadratic polynomial with zeroes 1/α, 1/β.

Solution:

α and β are the roots of the equation x^{2} + \(\sqrt{2}\)x + 3 = 0

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(x^{2} – 2x + 1) – 5x + 7 = 0

x^{2} (k) + x(-2k – 5) + k + 1 = 0

kx^{2} – 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

2(4k^{2} + 25 + 20k) = 9k (k + 7)

2(4k^{2} + 25 + 20k) = 9k^{2} + 63k

8k^{2} + 50 + 40k – 9k^{2} – 63k = 0

-k^{2} – 23k + 50 = 0

k^{2} + 23k – 5o = 0

(k + 25)(k – 2) = 0

k = -25 or 2

Question 5.

If the difference of the roots of the equation 2x^{2} – (a + 1)x + a – 1 = 0 is equal to their product then prove that a = 2.

Solution:

Question 6.

Find the condition that one of the roots of ax^{2} + 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

Question 7.

If the equations x^{2} – ax + b = 0 and x^{2} – ex + f = 0 have one root in common and if the second equation has equal roots that ae = 2(b + f).

Solution:

Question 8.

Discuss the nature of roots of

(i) -x^{2} + 3x + 1 = 0

(ii) 4x^{2} – x – 2 = 0

(iii) 9x^{2} + 5x = 0

Solution:

(i) -x^{2} + 3x + 1 = 0

x^{2} – 3x – 1 = 0 ———- (1)

Compare this equation with the equation

ax^{2} + bx + c = 0 ——– (2)

we have a = 1, b = -3, c = -1

Discriminant = b^{2} – 4ac

b^{2} – 4ac = (-3)^{2} – 4 × 1 × – 1

= 9 + 4 =13

b^{2} – 4ac = 13 > 0

∴ The two roots are real and distinct.

(ii) 4x^{2} – x – 2 = 0

4x^{2} – x – 2 = 0 ——(3)

Compare this equation with the equation

ax^{2} + bx + c = 0 (4)

we have a = 4 , b = – 1, c = – 2

Discriminant = b^{2} – 4ac

b^{2} – 4ac = (-1)^{2} – 4 (4) (-2)

= 1 + 32

= 33

b^{2} – 4ac = 33 >0

∴ The two roots are real and distinct.

(iii) 9x^{2} + 5x = 0

9x^{2} + 5x = 0 ——- (5)

Compare this equation with the equation

ax^{2} + bx + c = 0 ——– (6)

we have a = 9, b = 5 , c = 0

Discriminant = b^{2} – 4ac

b^{2} – 4ac = 5^{2} – 4 × 9 × 0

b^{2} – 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 = x^{2} + x + 2

(ii) y = x^{2} – 3x – 1

(iii) y = x^{2} + 6x + 9

Solution:

Completing the Square Calculator is a free online tool that displays the variable value for the quadratic equation using completing the square method.

Question 10.

Write f(x) = x^{2} + 5x + 4 in completed square form.

Solution:

### 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 x^{2} = 2x (1 + 3k) + 7(3 + 2k) = 0 has real and equal roots.

Solution:

The equation is x^{2} – x(2) (1 + 3k) + 7 (3 + 2k) = 0

The roots are real and equal

⇒ ∆ = 0 (i.e.,) b^{2} – 4ac = 0

Here a = 1, b = -2 (1 + 3k), c = 7(3 + 2k)

So b^{2} – 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 + 9k^{2} + 6k – 21 – 14k = 0

9k^{2} – 8k – 20 = 0

(k – 2)(9k + 10) = 0

To solve the quadratic inequalities ax^{2} + bx + c < 0 (or) ax^{2} + bx + c > 0

Question 2.

If the sum and product of the roots of the quadratic equation ax^{2} – 5x + c = 0 are both equal to 10 then find the values of a and c.

Solution:

The given equation is ax^{2} – 5x + c = 0

Let the roots be α and β Given α + β = 10 and αβ = 10

Question 3.

If α and β are the roots of the equation 3x^{2} – 4x + 1 = 0, form the equation whose roots are \(\frac{\alpha^{2}}{\beta}\) and \(\frac{\beta^{2}}{\alpha}\)

Solution:

Question 4.

If one root of the equation 3x^{2} + kx – 81 = 0 is the square of the other then find k.

Solution:

Question 5.

If one root of the equation 2x^{2} – ax + 64 = 0 is twice that of the other then find the value of a.

Solution: