Question

Find the value of p and q if they are the zeroes of 2x^2+px+q

Really urgent !!!!!!!!!!!

Answer 1

Answer:

2x2 - 5x - 3

2x2 - 6x +1x -3

2x (x - 3) +1 (x - 3)

(x - 3) (2x+1)

x = 3, x = -1/2

so, zeroes of x2 +px+q are 2*3 = 6 and 2 * -1/2 = -1

let the zeroes be a = 6 and b = -1

a+b = -coefficient of x/coefficient of x2

6 + (-1) = -p/1

5 = -p

p = -5

a*b = constant term/coefficient of x2

6 * (-1) = q/1

q = -6

Answer 2

Answer: Two possible cases are there : p = q = 0 and p = 1/2 , q = -3/4

Step-by-step explanation:

2x² + px + q = 0 , Zeroes ⇒ p, q

For a quadratic equation of form ax² + bx + c = 0, sum of roots is given in terms of coefficient by -b/a and product of roots is given by c/a

2x² + px + q = 0

Sum of roots = $\frac{-p}{2}$

⇒ $p + q = \frac{-p}{2}$

⇒ $q = \frac{-3p}{2}$

Product of roots =  $\frac{q}{2}$

⇒ $pq = \frac{q}{2}$

⇒ $q(p-\frac{1}{2}) = 0$

⇒ $\frac{-3p}{2} (p-\frac{1}{2}) = 0$

⇒ $p = 0 , \frac{1}{2}$

$q = \frac{-3p}{2}$

When p = 0,

q = 0

When p = 1/2

$q = \frac{-3}{4}$

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