Question

If alpha, beta are zeroes of the polynomial p(x)= x^2-3ax +a^2. Find the value of a if it is given that alpha^2+ beta^2=7/4

Answer 1

x^2 - 3ax + a^2 = 0

alpha + beta = 3a

alpha*beta= a^2

alpha^2 + beta^2 = 7/4

(alpha + beta)^2 - 2*alpha*beta = 7/4

9a^2 - 2a^2 = 7/4

7a^2 = 7/4

a^2 = 1/4

a= +1/2, -1/2

Answer 2

The value of a is  $\frac{1}{2} \ and \ -\frac{1}{2} .$

Step-by-step explanation:

Given:

$\alpha \ and \ \beta$ are zeroes of the polynomial $p(x)=x^{2} -3ax+a^{2}$.

$\alpha ^{2} +\beta ^{2} =\frac{7}{4}$.

To Find:

The value of a.

Solution:

As given-$\alpha \ and \ \beta$ are zeroes of the polynomial $p(x)=x^{2} -3ax+a^{2}$.

Since,α and β are the zeroes of the given polynomial.

Therefore,

$\alpha +\beta = - \frac{(-3a)}{1} =3a$

$\alpha .\beta =\frac{a^{2} }{1} =a^{2}$

As given- $\alpha ^{2} +\beta ^{2} =\frac{7}{4}$.

We know that

$(\alpha +\beta )^{2} =\alpha ^{2} +\beta ^{2} +2\alpha \beta$ --------- equation 01,

Putting the value of $(\alpha +\beta ),\alpha \beta \ and \ \alpha ^{2} +\beta ^{2}$ in equation no.01.

$(3a )^{2} =\frac{7}{4} +2a^{2}$

$9a^{2} -2a^{2} =\frac{7}{4}$

$7a^{2} =\frac{7}{4}$

$a^{2} =\frac{1}{4}$

$a= \frac{1}{2} \ and \ -\frac{1}{2} .$

Thus,  the value of a is $\frac{1}{2} \ and \ -\frac{1}{2} .$

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