Question

x square+kx-4=0 has - 4 as its root and x spare + Px + K =0 has its equal roots. find the value of p and k​

Answer 1

Step-by-step explanation:

Given :-

x²+kx-4=0 has - 4 as its root and x² + px + k =0 has equal roots.

To find :-

Find the values of p and k ?

Solution :-

Given quadratic equation is x²+kx-4 = 0

Let f(x) = x²+kx-4 = 0

Given root of f(x) = -4

If -4 is a root of the given equation then it satisfies the given equation

=> f(-4) = 0

=> (-4)²+k(-4)-4 =0

=> 16-4k-4 = 0

=> 12-4k = 0

=> 12 = 4k

=> 4k = 12

=> k = 12/4

=> k = 3

The value of k = 3

and

Given another equation x²+px+k = 0

Put k = 3 then it becomes x²+px+3 = 0

On comparing with the standard quadratic equation ax²+bx+c = 0

We have

a = 1

b = p

c = 3

It is given that it has equal roots

We know that

The quadratic equation ax²+bx+c = 0 has equal roots if the discriminant b²-4ac = 0

=> b²-4ac = 0

=>p²-4(1)(3) = 0

=> p² -12 = 0

=> p² = 12

=> p = ±√12

=> p = ±√(3×2×2)

=> p = ±2√3

The values of p = 2√3 and -2√3

Answer:-

The value of k = 3

The values of p = 2√3 and -2√3

Used formulae:-

• The standard quadratic equation ax²+bx+c = 0
• The quadratic equation ax²+bx+c = 0 has equal roots if the discriminant b²-4ac = 0

Points to know:-

• The nature of the roots of a quardratic equation is depending upon the value of its discriminant.
• If D= b²-4ac = 0 then it has two equal and real roots.
• If If D= b²-4ac < 0 then it has no real roots.