# If 3 is a root of the quadratic equation x2+ bx-15=0 determine the value of b. Find the value of the other root.

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According to the remainder theorem of polynomials, ‘a’ is a root of a function f(x) if and only if f(a)=0.

Here, `f(x)=x^2+bx-15`

3 is a root of it.

Put f(3) = 0 to get:

`3^2+ b*3-15=0`

`rArr 3b-6=0`

`rArr 3b=6`

`rArr b=2`

The given equation then simplifies to: `x^2+ 2x-15=0`.

The equation can also be written as: `x^2+ 5x-3x-15=0`

`rArr x(x+ 5)-3(x+5)=0`

`rArr (x+5)(x-3)=0`

**So, the other root is -5**.

**QUESTION:-**

**If 3 is a root of the quadratic equation x^2+ bx-15=0 determine the value of b. Find the value of the other root.**

**SOLUTION:-**

Now according to the question, 3 is a root of the quadratic equation given.

And according to the Remainder theorem of Polynomials,

‘a’ is a root of a function f(x) if and only if f(a)=0.

**With the help of this given root we will first find the value of b.**

Therefore,

`f(x)=ax^2+bx+c=0`

`f(3)=a(3)^2+b(3)+c=0`

Where;

according to the equation;

`x^2+bx-15=0`

a = 1

b = ?

c = 15

Insert values;

`f(3)=1(3)^2+b(3)-15=0`

`9+3b-15=0`

`3b-6=0`

`6=3b`

`3b=6`

`b=6/3`

`b=2`

**Now since we found out the value of b which is 2, now we will insert in the given equation and simplify it.**

**b = 2**

**The equation is: `f(x)=x^2+bx-15=0`**

Insert the value of b in the equation and then simplify;

`x^2+2x-15=0`

`x^2-3x+5x-15=0`

`x(x-3)+5(x-3)=0`

`(x+5)(x-3)=0`

`x+5=0,x-3=0`

`x=-5,x=3`

`SolutionSet:{-5,3}`

So the other root is -5.