Question

If alpha; beta are zeros of quadratic polynomial 2 x^ 2 + 5 x + k find the value of k such that alpha + beta whole square minus alpha beta =24

Answer 1

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Answer 2

Answer:

K= -71/2

Step-by-step explanation:

Given : P(X) :

$2x {}^{2} + 5x + k$

standard form :

.

$a {x}^{2} + bx + c$

( a= 2, b= 5, c= k)

sum of zeros

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

= -(5)/2 = -5/2

product of zeros

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

= k /2

Given:

$( \alpha + \beta ) { }^{2} - \alpha \beta = 24$

(-5/2)^2-k/2 = 24

25/4- k /2 = 24

lcm = 4

25-2 k /4= 24

25-2k = 24*4

25-2k = 96

2k = 96-25

2k = -71

K = -71/2