Math, asked by nikitayenarwar, 1 year ago

alpha 10 & beta 10 Q. From the quadratic equation from the roots given below

Answers

Answered by jaisurya64
1
 product of the roots = 0

⇒ x2 - Sx + P = 0, where S = sum of the roots and P = product of the roots ............... (i)

Formula (i) is used for the formation of a quadratic equation when its roots are given.

For example suppose we are to form the quadratic equation whose roots are 5 and (-2). By formula (i) we get the required equation as

x2 - [5 + (-2)]x + 5 ∙ (-2) = 0

⇒ x2 - (-3)x + (-10) = 0

⇒ x2 + 3x - 10 = 0


Solved examples to form the quadratic equation whose roots are given:

1. Form an equation whose roots are 2, and - 12.

Solution:

The given roots are 2 and -12.

Therefore, sum of the roots, S = 2 + (-12) = 32

And tghe product of the given roots, P = 2 ∙ -12= - 1.

Therefore, the required equation is x2 – Sx + p

i.e., x2 - (sum of the roots)x + product of the roots = 0

i.e., x2 - 32x – 1 = 0

i.e, 2x2 - 3x - 2 = 0


2. Find the quadratic equation with rational coefficients which has 13+2√2 as a root.

Solution:

According to the problem, coefficients of the required quadratic equation are rational and its one root is 13+2√2 = 13+2√2 ∙ 3−2√23−2√2 = 3−2√29−8= 3 - 2√2.

We know in a quadratic with rational coefficients irrational roots occur in conjugate pairs).

Since equation has rational coefficients, the other root is 3 + 2√2.

Now, the sum of the roots of the given equation S = (3 - 2√2) + (3 + 2√2) = 6

Product of the roots, P = (3 - 2√2)(3 + 2√2) = 32 - (2√2)2 = 9 - 8 = 1

Hence, the required equation is x2 - Sx + P = 0 i.e., x2 - 6x + 1 = 0.


2. Find the quadratic equation with real coefficients which has -2 + i as a root (i = √-1).

Solution:

According to the problem, coefficients of the required quadratic equation are real and its one root is -2 + i.

We know in a quadratic with real coefficients imaginary roots occur in conjugate pairs).

Since equation has rational coefficients, the other root is -2 - i

Now, the sum of the roots of the given equation S = (-2 + i) + (-2 - i) = -4

Product of the roots, P = (-2 + i)(-2 - i) = (-2)2 - i2 = 4 - (-1) = 4 + 1 = 5

Hence, the required equation is x2 - Sx + P = 0 i.e., x2 - 4x + 5 = 0.



 





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