Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.4 Text Book Back Questions and Answers, Notes.

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

Question 1.

Construct a quadratic equation with roots 7 and – 3.

Answer:

The given roots are 7 and -3

Let α = 7 and β = -3

α + β = 7 – 3 = 4

αβ = (7)(-3) = -21

The quadratic equation with roots α and β is x^{2} – (α + β) x + αβ = 0

So the required quadratic equation is

x^{2} – (4) x + (-21) = 0

(i.e.,) x^{2} – 4x – 21 = 0

Question 2.

A quadratic polynomial has one of its zero 1 + √5 and it satisfies p(1) = 2. Find the quadratic polynomial.

Answer:

Let p(x) = ax^{2} + bx + c be the required quadratic polynomial.

Given p (1) = 2 , we have

a × 1^{2} + b × 1 + c = 2

a + b + c = 2 ——— (1)

Also given 1 + √5 is a zero of p(x)

∴ a(1 + √5)^{2} + b (1 + √5) + c = 0

a( 1 + 5 + 2√5) + b (1 + √5) + c = 0

6a + 2a√5 + b + b√5 + c = 0 ——— (2)

If 1 + √5 is zero then 1 – √5 is also a zero of p (x).

∴ a(1 – √5)^{2} + b (1 – √5) + c = 0

a( 1 – 2√5 + 5) + b (1 – √5) + c = 0

6a – 2a√5 + b – b√5 + c = 0 ——— (3)

Substituting the value of a in equation (4)

5 × – \(\frac{2}{5}\) + 2 × – \(\frac{2}{5}\) × √5 + b√5 = – 2

– 2 – \(\frac{4}{5}\)√5 + b√5 = – 2

b√5 = – 2 + 2 + \(\frac{4}{5}\) . √5

b√5 = \(\frac{4}{5}\) . √5

b = \(\frac{4}{5}\)

Substituting the value of a and b in equation (1), we have

∴ The required quadratic polynomial is

p(x) = \(-\frac{2}{5}\)x^{2} + \(\frac{4}{5}\)x + \(\frac{8}{5}\)

p(x) = \(-\frac{2}{5}\)(x^{2} – 2x – 4)

Question 3.

If α and β are the roots of the quadratic equation x^{2} + √2x + 3 = 0 form a quadratic polynomial with zeros \(\frac{1}{\alpha}, \frac{1}{\beta}\).

Answer:

Given α and β are the roots of the quadratic polynomial

x^{2} + √2x + 3 = 0 ——— (1)

∴ The required quadratic equation whose roots are \(\frac{1}{\alpha}, \frac{1}{\beta}\) is

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

Question 4.

If one root of k (x – 1)^{2} = 5x – 7 is double the other root, show that k = 2 or – 25

Answer:

The given quadratic equation is

k(x – 1)^{2} = 5x – 7

k(x^{2} – 2x + 1) – 5x + 7 = 0

kx^{2} – 2kx + k – 5x + 7 = 0

kx^{2} – (2k + 5)x + k + 7 = 0 ———- (1)

Let the roots be α and 2α

Sum of the roots α + 2α = –\(\frac{b}{a}\)

Product of te roots α(2α) = \(\frac{c}{a}\)

Using equation (2) and (3) we have

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

8k^{2} + 40k + 50 = 9k^{2} + 63k

9k^{2} + 63k – 8k^{2} – 40k – 50 = 0

k^{2} + 23k – 50 = 0

k^{2} + 25k – 2k – 50 = 0

k(k + 25) – 2(k + 25) = 0

(k – 2) (k + 25) = 0

k – 2 = 0 or k + 25 = 0

k = 2 or k = – 25

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.

Answer:

The given quadratic equation is

2x^{2} – (a + 1) x + a – 1 = 0 ———– (1)

Let α and β be the roots of the given equation

Given that α – β = αβ —— (2)

Sum of the roots α + β = – \(\frac{b}{a}\)

Substituting the values of α and β in equation (2)

2(a – 1) = a

2a – 2 – a = 0

a – 2 = 0

⇒ a = 2

Question 6.

Find the condition that one of the roots of ax^{2} + bx + c may be

(a) negative of the other

(b) thrice the other

(c) reciprocal of the other.

Answer:

The given quadratic equation is

ax^{2} + bx + c = 0 ——- (1)

Let α and β be the roots of the equation (1) then

Sum of the roots α + β = ——- (2)

Product of the roots αβ = ——- (3)

(a) Given one root is the negative of the other

β = – α

(2) ⇒ α + (-α) = – \(\frac{b}{a}\)

0 = – \(\frac{b}{a}\)

⇒ b = 0

(3) ⇒ α(-α) = \(\frac{c}{a}\)

– α^{2} = \(\frac{c}{a}\)

Hence the required condition is b = 0

(b) Given that one root is thrice the other

β = 3α

When is the required condition?

(c) One root is reciprocal of the other

When is the required condition?

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 then prove that ae = 2(b + f).

Answer:

The given quadratic equations are

x^{2} – ax + b = 0 ———- (1)

x^{2} – ex + f = 0 ——— (2)

Let α be the common root of the given quadratic equations (1) and (2)

Let α, β be the roots of x^{2} – ax + b = 0

Sum of the roots α + β = \(-\left(-\frac{a}{1}\right)\)

α + β = a ———- (3)

Product of the roots αβ = \(\frac{b}{1}\)

αβ = b ——– (4)

Given that the second equation has equal roots.

∴ The roots of the second equation are a, a

Sum of the roots α + α = \(-\left(-\frac{e}{1}\right)\)

2α = e ——— (5)

Product of the roots α.α = \(\frac{f}{1}\)

α^{2} = f ———- (6)

ae = (α + β)2α (Multiplying equations (3) and (5))

ae = 2α2 + 2αβ

ae= 2 (f) + 2b From equations (4) and (6)

ae= 2(f + b) Hence proved.

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.

Answer:

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

⇒ comparing with ax^{2} + bx + c = 0

∆ = b^{2} – 4ac = (3)^{2} – 4(1)(-1) = 9 + 4 = 13 > 0

⇒ The roots are real and distinct

(ii) 4x^{2} – x – 2 = 0

a = 4, b = -1, c = -2

∆ = b^{2} – 4ac = (-1)^{2} – 4(4)(-2) = 1 + 32 = 33 >0

⇒ The roots are real and distinct

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

a = 9, b = 5, c = 0

∆ = b^{2} – 4ac = 5^{2} – 4(9)(0) = 25 > 0

⇒ The 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 – 7

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

Answer:

(i) y = x^{2} + x + 2

y = x^{2} + x + 2 ——— (1)

Compare this equation with the equation

ax^{2} + bx + c = 0

we have a = 1 , b = 1, c = 2

b^{2} – 4ac = 1^{2} – 4 × 1 × 2 = 1 – 8

b^{2} – 4ac = – 7 < 0

Since the discriminant is negative the quadratic equation has no real roots and therfore the graph does not meet x-axis.

(ii) y = x^{2} – 3x – 7

y = x^{2} – 3x – 7 ——— (2)

Compare this equation with the equation ax^{2} + bx + c = 0

we have a = 1 , b = – 3 , c = – 1

b^{2} – 4ac = (-3)^{2} – 4(1)(-1)

= 9 + 4

b^{2} – 4ac = 13 > 0

Since the discriminant is positive the quadratic equation has real and distinct roots and therefore the graph intersect the x – axis at two different points,

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

y = x^{2} + 6x + 9 ——— (3)

Compare this equation with the equation

ax^{2} + bx + c = 0

we have a = 1 , b = 6, c = 9

b^{2} – 4ac = 6^{2} – 4 × 1 × 9

= 36 – 36 =0

Since the discriminant is zero the quadratic equation has real and equal roots and therefore the graph touches the x-axis at one point.

Question 10.

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

Answer:

The given quadratic equation is

f(x) = x^{2} + 5x + 4

Let y = x^{2} + 5x + 4

y – 4 = x^{2} + 5x