Question

If one of the roots of quadratic equation X2 – kX + 27 = 0 is 3 then find the value of ‘k’.
A) 10
B) 12
C) -12
D) 16​

Answer 1

Answer:

Concept:

For any quadratic equation, ax2 + bx + c = 0. We have discriminant, D = b2 - 4ac, then the given quadratic equation has:

I. Distinct and real roots if D > 0.

II. Real and repeated roots, if D = 0.

III. Complex roots and conjugate of each other, D < 0.

Calculation:

Given: x2 + mx + 2 = 0

By comparing the given equation with the standard quadratic equation ax2 + bx + c = 0, we get a = 1, b = m and c = 2.

⇒ D = b2 - 4ac = m2 - 8

For real roots, D ≥ 0

⇒ D = m2 - 8 ≥ 0 ⇒ m2 ≥ 8.

Answer 2

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Answer:

B) 12

Step-by-step explanation:

roots are - a and b

a=3 (given)

quadratic equation is of form X² + (a+b)x + ab

here p(x) = x²+Kx+27

so

(a+b) = k

ab= 27

now,

ab= 27

3×b=27

b= 27/3

b= 9

k = a + b

k = 3+9

k = 12

so option B) 12 is correct