# If x^2 - bx + 4 = 0 has common roots, what is the value of b

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The equation x^2 - bx + 4 = 0 has common roots.

The roots of the quadratic equation ax^2 + bx + c = 0 are common if b^2 = 4ac

For the equation x^2 - bx + 4 = 0, a = 1 and c = 4

`b^2 = 4ac`

=> `b^2 = 4*1*4`

=> `b^2 = 16`

=> `b = +- 4`

**The value of b can be **`+- 4`

To help you gain better insight into the question, take a look at the quadratic curves with various values of b (or -b for your case)

Cheers.

`x^2 - bx + 4 = 0` has common roots, then this is factorable.

Therefore: `(x - 2) (x - 2) = 0`

Then this means umultiplied this gives us:

`x^2 - 4x + 4 = 0`

**Hence, b = 4**

`x^2-bx+4=0`

`x^2-2xx(b/2)x+(b/2)^2=(b/2)^2-4`

`(x-b/2)^2=(+-sqrt((b^2-16)/4))^2`

Equation has common roots if

`(b^2-16)/4=0`

`b^2-16=0`

`b^2=16`

`b^2=(+-4)^2`

`b=+-4`

Thus if we substitute `b=+-4` , we have common roots.

`x^2 - bx + 4 `

take the square root of a and c to factor

`(x+2)(x+2) `

unfoil

`x xx x=x^2 `

`x xx 2=2x `

`2xx x =2x`

`2 xx 2 =4`

you end up with

`x^2+2x+2x+4 `

combine like terms

`x^2+4x+4 `

therefore b is 4