Question

If a and B are the zeroes of the quadratic polynomial p(x)=3x²-4x-7, then find the
quadratic polynomial whose zeroes are 1/a and 1/b​

Answer 1

Answer:

If a quadratic equation is of the form: ax² + bx + c, then the relation betwen zeros is given as:

⇒ Sum of Zeros = -b/a

⇒ Product of Zeros = c/a

According to the given question, the zeros are 'a' and 'b', the quadratic polynomial is: p(x) = 3x² - 4x - 7

From this we get:

• a = 3
• b = -4
• c = -7

Therefore the sum of zeros and product of zeros are given as:

⇒ a + b = 4/3 ...(1)

⇒ ab = -7/3 ...(2)

Now, the relation between zeros and the quadratic equation is of the form:

⇒ Quadratic Equation = x² - (Sum of zeros) x + (Product of zeros)

According to the question, the new zeros are: 1/a and 1/b.

Hence the sum of zeros is:

⇒ 1/a + 1/b = (a+b)/ab

Substituting the values of (a+b) and (ab) from (1) and (2) we get:

⇒ 1/a + 1/b = (4/3) / (-7/3)

⇒ 1/a + 1/b = -4/7

And product of the zeros are:

⇒ 1/a × 1/b = 1/ab

⇒ 1/a × 1/b = 1 / (-7/3)

⇒ 1/a × 1/b = -3/7

Hence forming the new quadratic equation we get:

⇒ x² - ( -4/7 ) x - 3/7 = 0

⇒ x² + 4x/7 - 3/7 = 0

Taking LCM we get:

⇒ 7x² + 4x - 3 = 0

Hence the required quadratic polynomial with zeros (1/a) and (1/b) is:

• 7x² + 4x - 3

Answer 2

7x²+4x-3

Step-by-step explanation:

Solution:

3x²- 4x-7 = 0

On factorising,

• => 3x²+3x-7x-7 =0

• => 3x(x+1)-7(x+1) = 0

• => (3x-7)(x+1) = 0

• => 3x-7 = 0 and x+1 = 0

• => x = 7/3 and -1

Hence, 7/3 and -1 are the zeroes of the polynomial.

Let:

• a = 7/3

• b = -1

Then,

• 1/a = 3/7

• 1/b = -1

• => Sum of zeroes = 3/7+(-1) = -4/7 = -b/a

• => Product of zeroes = 3/7×(-1) = -3/7 = c/a

Formula to find Quadratic polynomial:

• => k[x²-(Sum of zeroes)x + (Product of zeroes)]

Note: Here, k is any real number.

• => k[x²+4/7x-3/7]

To remove the fraction, We are going to multiply the whole polynomial by 7.

Let k be 7:

• => 7[x²+4/7x-3/7]

• Required polynomial: => 7x²+4x-3