Question

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

Answer 1

Answer:

$\large\boxed{a=\dfrac{225}{16}}$

$\dfrac{225}{16}x^2+15x+4=\left(\pm\dfrac{15}{4}x\pm2\right)^2}$

Step-by-step explanation:

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

Answer 2

Answer:

$\frac{225}{16}$

Step-by-step 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:

$m=\frac{15}{2}$

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

$(\frac{15}{2})^2=4a$

$\frac{225}{4}=4a$

Divide both sides by 4:

$\frac{225}{16}=a$

So $\frac{225}{16}x^2+15x+4=(\frac{15}{4}x+2)^2$.

Let's check:

$(\frac{15}{4}x+2)^2$

$(\frac{15}{4}x+2)(\frac{15}{4}x+2)$

$\frac{15}{4}x \cdot \frac{15}{4}x+\frac{15}{4}x \cdot 2+2\cdot \frac{15}{4}x+2\cdot 2$

$\frac{225}{16}x^2+\frac{15}{2}x+\frac{15}{2}x+4$

$\frac{225}{16}x^2+\frac{30}{2}x+4$

$\frac{225}{16}x^2+15x+4$