Question

one root of x²+px+16=0 is four times the other. Find possible values of p

Answer 1

Answer :

p = ± 10

Note:

★ The possible values of the variable which satisfy the equation are called its roots or solutions .

★ A quadratic equation can have atmost two roots .

★ The general form of a quadratic equation is given as ; ax² + bx + c = 0

★ If α and ß are the roots of the quadratic equation ax² + bx + c = 0 , then ;

• Sum of roots , (α + ß) = -b/a

• Product of roots , (αß) = c/a

Solution :

Here ,

The given quadratic equation is ;

x² + px + 16 = 0

Now ,

Comparing the given quadratic equation with the general quadratic equation , ax² + bx + c = 0 , we have ;

a = 1

b = p

c = 16

Also ,

It is given that , one root of the given quadratic equation is four times the other .

Thus ,

Let α and 4α be the roots of the given quadratic equation .

Now ,

The sum of roots of the given quadratic equation will be given as ;

=> α + 4α = -b/a

=> 5α = -p/1

=> p = -5α -----(1)

Also ,

Product of roots will be given as ;

=> α•4α = c/a

=> 4α² = 16/1

=> α² = 16/4

=> α² = 4

=> α = √4

=> α = ±2

• If α = 2 , then using eq-(1) , we have ;

=> p = -5α

=> p = -5•2

=> p = -10

• If α = -2 , then using eq-(1) , we have ;

=> p = -5α

=> p = -5•(-2)

=> p = 10

Hence , p = ± 10 .