Question

If α and β are zeroes of 2x²+3x-6,then α²+β²is

9/4

25/4

33/4

-3/2

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Answer 1

Answer:

33/4

Step-by-step explanation:

2(x² + $\frac{3}{2}$x - 3) = 2[(x² + 2×$\frac{3}{4}$x + $\frac{9}{16}$) - $\frac{9}{16}$ - 3 =

2[(x + $\frac{3}{4}$)² - $\frac{57}{16}$] = 2(x + $\frac{3}{4}$ - (√57)/4)(x + $\frac{3}{4}$ + (√57)/4)

α = (√57 - 3) / 4

β = - (√57 + 3) / 4

α² + β² = [( (√57 - 3)² +  (√57 + 3)²] / 16 =

(57 - 6√57 + 9 + 57 + 6√57 + 9) / 16

(114 + 18) / 16 = 132 / 16 = 33/4

Answer 2

Answer:

33/4

Step-by-step explanation:

$\alpha \beta = \frac{c}{a} \\ \alpha \beta = \frac{ - 6}{2} \\ \alpha \beta = - 3 \: and\\ \alpha + \beta = - \frac{ b}{a} \\ \alpha + \beta = - \frac{3 }{2} \\ ( { \alpha + \beta )}^{2} = \frac{9}{4} \\ { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta = \frac{9}{4} \\ { \alpha}^{2} + { \beta }^{2} - 2 \times 3 = \frac{9}{4} \\ { \alpha}^{2} + { \beta }^{2} = \frac{9}{4} + 6 \\ { \alpha }^{2} + { \beta }^{2} = \frac{33}{4}$