Question

Find $a$ such that $ax^2+15x+4$ is the square of a binomial

Answer 1

`(225)/(16)`

Explanation:

`ax^2+15x+4`

We want to find
`a` such that:

two same numbers,
`m \text{ and } m`, multiply to be
`4a` and

add up to be 15.

`m^2=4a`

`2m=15`

We can solve the last equation by dividing both sides by 2:

`m=(15)/(2)`

So this means
`m^2=4a` can be rewritten as:

`((15)/(2))^2=4a`

`(225)/(4)=4a`

Divide both sides by 4:

`(225)/(16)=a`

So
`(225)/(16)x^2+15x+4=((15)/(4)x+2)^2`.

Let's check:

`((15)/(4)x+2)^2`

`((15)/(4)x+2)((15)/(4)x+2)`

`(15)/(4)x \cdot (15)/(4)x+(15)/(4)x \cdot 2+2\cdot (15)/(4)x+2\cdot 2`

`(225)/(16)x^2+(15)/(2)x+(15)/(2)x+4`

`(225)/(16)x^2+(30)/(2)x+4`

`(225)/(16)x^2+15x+4`

Answer 2

`\large\boxed{a=(225)/(16)}`

`(225)/(16)x^2+15x+4=\left(\pm(15)/(4)x\pm2\right)^2}`

Explanation:

`ax^2+15x+4=(bx+c)^2\qquad\text{use}\ (x+y)^2=x^2+2xy+y^2\\\\ax^2+15x+4=(bx)^2+2(bx)(c)+c^2\\\\ax^2+15x+4=b^2x^2+2bcx+c^2\\\\\text{Therefore}\\\\(1)\qquad b^2=a\\(2)\qquad2bc=15\\(3)\qquad c^2=4\\\\\text{From}\ (3):\\\\c^2=4\to c=\pm\sqrt4\to c=\pm2\\\\\text{Substitute to (2):}\\\\2b(\pm2)=15\\\pm4b=15\qquad\text{divide both sides by}\ \pm4\\\\b=\pm(15)/(4)\\\\\text{Substitute to (1):}\\\\a=\left(\pm(15)/(4)\right)^2=(225)/(16)`