Question

if alpha, beta are the roots of 3x^2-5x+7=0 then find alpha square plus beta square​

Answer 1

Answer:

The given equation is:

                                    3x^2-5x+7 =0

here,

       a= 3    , b= -5  , c=7

Step-by-step explanation:

Alpha= A

Beta= B

so,

    Sum of the roots = A+B= -b/a = 5/3

  Product of the roots=AB =c/a= 7/3

now,

A^2 + B^2= ?

=(A+B)^2-2AB .......(1)        {when we solve this :

                                      = A^2+B^2+2AB-2AB

                                      = A^2+B^2  }

now putting the value   ALPHA+BETA (A+B) and (ALPHA)(BETA) in eq (1)

so,

=(5/3)^2-2(7/3)

=25/9 - 14/3

= (25-42)/9

= -17/9

so,-17/9 is the required answer.

Answer 2

α²+β² = -17/9.

Given:

The quadratic equation 3x²-5x+7=0.

To Find:

If α and β beta are the roots of 3x²-5x+7=0, then find (α²+β²).

Solution:

A quadratic equation is an equation having general form ax²+bx+c=0. It has a degree of 2.

Since a polynomial of 'n' degree has 'n' roots/zeros, a quadratic equation has 2 roots/zeros.

We are given that α and β beta are the roots of 3x²-5x+7=0.

Comparing this quadratic equation to its general form, we have:

a = the coefficient of x² = 3.

b = the coefficient of x = -5.

c = the constant term = 7.

The sum of roots = (α + β) = -b/a = -(-5)/3 = 5/3.

The product of roots = (α.β) = c/a = 7/3.

We know that,

(α + β)² = α²+2αβ+β².

⇒ α²+β² = (α + β)² - 2αβ.

Substituting the values of (α + β) and (α.β) into the above equation, we have:  

⇒ α²+β² = (5/3)² - 2(7/3) = 25/9 - 14/3 = -17/9.

Hence, α²+β² = -17/9.

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