FORMATION OF THE QUADRATIC EQUATION WHOSE ROOTS ARE GIVEN

Problem 1 :

Find the quadratic polynomial whose zeroes are −2 & 3

Solution :

Here α = -2 and β = 3

x² - (α + β) x + α β = 0

α + β = -2 + 3 ==> 1

α β = -2(3) ==> -6

x² - 1x + (-6) = 0

x² - x - 6 = 0

So, the required polynomial is x² - x - 6 = 0.

Problem 2 :

Find a quadratic polynomial whose zeroes are 2 + √3 and 2 – √3

Solution :

Here α = 2 + √3 and β = 2 – √3

α + β = 2 + √3 + 2 – √3 ==> 4

α β = (2 + √3)(2 – √3)

= 2² - √3²

= 4 - 3

= 1

x² - 4x + 1 = 0

So, the required polynomial is x² - 4x + 1 = 0.

Problem 3 :

Frame a quadratic polynomial p(x) whose sum of zeroes is -3 and the product of zeroes is -2/3.

Solution :

x² - (α + β) x + α β = 0

α + β = -3, αβ = -2/3

x² - (-3) x + (-2/3) = 0

x² + 3 x - 2/3 = 0

3x² +x - 2 = 0

Problem 4 :

Find a quadratic polynomial whose sum of zeroes is -12 and product is 14.

Solution :

x² - (α + β) x + α β = 0

α + β = -12, αβ = 14

x² - (-12) x + 14 = 0

x² + 12 x + 14 = 0

Problem 5 :

If the sum of the zeroes of the quadratic polynomial

f(x) = kx² - 2x + 3 is 3

find k.

Solution :

f(x) = kx² - 2x + 3

Creating this equation with sum and product of roots, we get

f(x) = k [x² - (2/k)x + (3/k) ]

α + β = 3

-2/k = 3

k = -6

Problem 6 :

The sum of the roots of the equation

3x² + kx + 5 = 0

will be equal to the product of its roots.

Solution :

3x² + kx + 5 = 0

Let create the quadratic equation in the form

x² - (α + β) x + α β = 0

3 [x² - (-k/3)x + (5/3)] = 0

x² - (-k/3)x + (5/3) = 0

α + β = -k/3 ------(1)

α β = 5/3 ------(2)

-k/3 = 5/3

k = -5

Problem 7 :

Find the value of k, given that the product of the roots of the equation

(k + 1)x² + (4k + 3)x + (k – 1) = 0 is 7/2

Solution :

By dividing (k + 1)x² + (4k + 3)x + (k – 1) = 0 by (k + 1), we get

x² + [(4k + 3)/(k + 1)]x + (k – 1)/(k + 1) = 0

Product of the roots = 7/2

(k – 1)/(k + 1) = 7/2

2(k - 1) = 7(k + 1)

2k - 2 = 7k + 7

2k - 7k = 7 + 2

-5k = 9

k = -9/5

Problem 8 :

Form an quadratic polynomial whose roots are 2, and -1/2.

Solution :

α + β = 2, α β = -1/2

By applying the values in this form, we get

x² - (α + β) x + α β = 0

x² - 2x + (-1/2) = 0

2x² - 4x - 1 = 0

Problem 9 :

What is an equation whose roots are 5+√2 and 5−√2

Solution :

α = 5+√2, β = 5-√2

α + β = 5 + √2 + 5 - √2

α + β = 10

αβ = (5+√2) (5-√2)

αβ = 5²-√2²

 = 25 - 2

= 23

x² - (α + β) x + α β = 0

x² - 10x + 23 = 0

Problem 10 :

Write down the quadratic equation in general form for which sum and product of the roots are given below.

– 7/2 , 5/2

Solution :

Sum of roots = -7/2 and product of roots = 5/2

x² - (α + β) x + α β = 0

x² - (-7/2) x + (5/2) = 0

2x² + 7x + 5 = 0