Question

Find the values of p and q for which 3/2 and -3 are two roots of the quadratic
equation px^2+qx-9=0

Answer 1

Answer:

p = 2

q = 3

Given a quadratic equation,

▶ px² + qx - 9 = 0

We have to find the values of p and q such that 3/2 and -3 are roots of the given quadratic equation.

Let the roots be 3/2 and -3 as given in the question.

∴ Sum of roots = 3/2 + (-3) = (3 - 6)/2 = -3/2 ...(1)

Similarly,

⇒ Product of roots = 3/2 × -3 = -9/2 ...(2)

Regarding the quadratic equation,

• px² + qx - 9 = 0

⇒ Sum of roots = -(coefficient of x) / (coefficient of x²)

⇒ - q / p = -3/2 [ from (1) ]

⇒ q = 3p/2 ...(3)

Also,

⇒ Product of roots = (constant term) / (coefficient of x²)

⇒ -9/2 = -9/p [ from (2) ]

⇒ 1/2 = 1/p

⇒ p = 2

Now, Substitute p = 2 in (3),

⇒ q = 3×2/2

⇒ q = 3

Hence, The values of p and q are 2 and 3 respectively.

Answer 2

Answer:

i

Step-by-step explanation:

quadratic equation= px²+ qx - 9 = 0.

values of p and q given as :

3/2 and -3

let the root be 3/2 and -3 given in the question

{sum of roots = 3/2+(-3) = (3-6)/2

= -3/2. let equation( 1 )

now ,

product of roots 3/2 *-3 = -9/2 let equation (2)

taking the quadratic equation,

px²+qx-9 = 0

sum of roots =- (coffecient of x) /( coffecient of x²)

= q= 3p/2 ( equation 3) .

product of roots = constant term /

coffecient of x²

p= 2

now substituting p= 2 in equation 3

q= 3.

hence the values are { p= 2 }

{q= 3} .