Question

1. Show that x2 + 4x + 7 has no zero.

2. If ½ is the zero of the p(x) = 2x4 –ax3 + 4x2 +2x + 1, find the value of a .

3. If 1 and -1 are zeroes of p(x) = ax3 +x2 -2x + b , find a and b.

4. If 2 and ½ are zeroes of p(x) = ax2 + 5x + b , show that a = b​

Answer 1

Answer :

Question - 1 :

The given quadratic equation is x² + 4x + 7

Let p(x) = x² + 4x + 7

Factorizing it,

p(x) = x² + 4x + 7

p(x) = x² + 4x + 4 + 3

p(x) = [x² + 2(x)(2) + 2²] + 3

p(x) = (x + 2)² + 3

For any value of x, (x + 2)² will be positive.

So, the p(x) will never be zero.

Hence, it has no zero.

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Question - 2 :

If 'a' is a zero of the polynomial p(x), then p(a) = 0.

Given polynomial, p(x) = 2x⁴ – ax³ + 4x² + 2x + 1

Put x = 1/2,

p(1/2) = 0

2(1/2)⁴ – a(1/2)³ + 4(1/2)² + 2(1/2) + 1 = 0

2(1/16) – a(1/8) + 4(1/4) + 1 + 1 = 0

1/8 – a/8 + 1 + 1 + 1 = 0

1/8 – a/8 + 3 = 0

(1 – a)/8 = –3

1 – a = (-3) × 8

1 – a = –24

a = 1 + 24

a = 25

Therefore, the value of a is 25

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Question - 3 :

Given cubic polynomial, p(x) = ax³ + x² – 2x + b

1 and -1 are the zeroes.

Put x = 1,

p(1) = 0

a(1)³ + 1² - 2(1) + b = 0

a(1) + 1 - 2 + b = 0

a - 1 + b = 0

a + b = 1

b = 1 – a ––> eqn. 1

Put x = -1,

p(-1) = 0

a(-1)³ + (-1)² - 2(-1) + b = 0

a(-1) + 1 + 2 + b = 0

-a + 3 + b = 0

-a + b = –3

–a + 1 – a = –3 [eqn. 1[

–2a + 1 = –3

–2a = –3 – 1

–2a = –4

2a = 4

a = 2

Put a = 2 in eqn. 1,

b = 1 – 2

b = –1

Therefore, a = 2 and b = –1

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Question - 4 :

2 and ½ are zeroes of p(x) = ax² + 5x + b

From the relation between zeroes and coefficients of quadratic polynomial,

Product of zeroes = constant/x² coefficient

2 × ½ = b/a

1 = b/a

b = a

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Answer 2

Question 1) :- Show that x² + 4x + 7 has no real values of zero.

Solution :-

when know that, when

• D(discriminant) < 0 , the quadratic equation has no real roots.
• D = b² - 4ac

so, in x² + 4x + 7 we have,

• a = 1
• b = 4
• c = 7

then,

→ D = b² - 4ac

→ D = (4)² - 4*1*7

→ D = 16 - 28

→ D = (-12)

since,

→ (-12) < 0 .

→ D < 0 .

therefore, the given quadratic equation has no real value of zeros .

Question 2) :- ½ is the zero of the p(x) = 2x⁴ –ax³ + 4x² +2x + 1, find the value of a .

Solution :-

since (1/2) is the zero of p(x) , it gives remainder as 0 .

so,

→ p(x) = 2x⁴ –ax³ + 4x² +2x + 1

→ p(1/2) = 0

→ 2(1/2)⁴ - a(1/2)³ + 4(1/2)² + 2(1/2) + 1 = 0

→ 2(1/16) - (a/8) + 4(1/4) + 1 + 1 = 0

→ (1/8) - (a/8) + 3 = 0

→ (a/8) = (25/8)

→ a = 25 (Ans.)

Question 3) :- 1 and -1 are zeroes of p(x) = ax³ +x² -2x + b , find a and b.

putting x = 1 ,

→ p(x) = ax³ + x² - 2x + b

→ p(1) = 0

→ a(1)³ + (1)² - 2(1) + b = 0

→ a + 1 - 2 + b = 0

→ a + b = 1 -------- Eqn.(1)

putting x = (-1),

→ → p(x) = ax³ + x² - 2x + b

→ p(-1) = 0

→ a(-1)³ + (-1)² - 2(-1) + b = 0

→ -a + 1 + 2 + b = 0

→ b - a = -3 -------- Eqn.(2)

adding Eqn.(1) and Eqn.(2),

→ a + b + b - a = 1 - 3

→ 2b = (-2)

→ b = (-1) (Ans.)

putting value of b in Eqn.(1),

→ a - 1 = 1

→ a = 2 (Ans.)

Question 4) :- . If 2 and ½ are zeroes of p(x) = ax² + 5x + b , show that a = b

Solution :-

we know that, for quadratic equation ax² + bx + c = 0,

• Product of zeros = c/a .

comparing ax² + bx + c = 0 by ax² + 5x + b, we get,

• a = a
• b = 5
• c = b

given zeroes are 2 and (1/2) .

so,

→ 2 * (1/2) = c/a

→ 1 = b/a

→ b = a .

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