Question

if alpha and beta are the zeroes of 2x^2 - 3x + 7 find alpha square plus beta square​

Answer 1

EXPLANATION.

α, β are the zeroes of the polynomial.

⇒ 2x² - 3x + 7.

As we know that,

Sum of the zeroes of the quadratic equation.

⇒ α + β = -b/a.

⇒ α + β = -(-3)/2 = 3/2.

Products of the zeroes of the quadratic equation.

⇒ αβ = c/a.

⇒ αβ = 7/2.

To find :

⇒ α² + β².

As we know that,

Formula of :

⇒ (x² + y²) = (x + y)² - 2xy.

Using this formula in the equation, we get.

⇒ α² + β² = (α + β)² - 2αβ.

Put the values in the equation, we get.

⇒ α² + β² = (3/2)² - 2(7/2).

⇒ α² + β² = 9/4 - 7.

⇒ α² + β² = (9 - 28)/7.

⇒ α² + β² = - 19/7.

                                                                                                                     

MORE INFORMATION.

Conjugate roots.

(1) = If D < 0.

One roots = α + iβ.

Other roots = α - iβ.

(2) = If D > 0.

One roots = α + √β.

Other roots = α - √β.

Answer 2

$\huge \boxed{ \sf \red{ { \alpha }^{2} + { \beta }^{2} = \frac{ - 19}{4} }}$

Step-by-step explanation:

2x² - 3x + 7

.

It is in the standard form of quadratic equation (ax² + bx + c), here

a = 2

b = -3

c = 7

.

Sum of zeroes,

$\alpha + \beta = - \frac{b}{a} \\ = - \frac{ - 3}{2} \\ = \frac{3}{2}$

product of zeroes,

$\alpha \beta = \frac{c}{a} \\ = \frac{7}{2}$

Now,

${ \alpha }^{2} + { \beta }^{2} = ( { \alpha + \beta) }^{2} - 2 \alpha \beta \\ \\ = {( \frac{3}{2}) }^{2} - 2( \frac{7}{2} ) \\ \\ = \frac{9}{4} - \frac{14}{2} \\ \\ = \frac{9 - 28}{4} \\ \\ = \frac{ - 19}{4}$

Hence alpha² + beta² = -19/4.

hope it helps.