Question

the equation -2x^2+3=-9x is rewritten in the form of -2(x-p)^2 +q=0. What is the value of q?the equation -2x^2+3=-9x is rewritten in the form of -2(x-p)^2 +q=0. What is the value of q?

Answer 1

Value of q is $\frac{105}{8}$ .

Step-by-step explanation:

We have , the equation $-2x^2+3=-9x$ is rewritten in the form of $-2(x-p)^2 +q=0$ . We need to find What is the value of q . Let's find out:

Let's simplify equation $-2x^2+3=-9x$ into $-2(x-p)^2 +q=0$ :

⇒ $-2x^2+3=-9x$

⇒ $-2x^2+9x+3=0$

⇒ $2x^2-9x=3$

⇒ $2(x^2-\frac{9}{2}x)=3$

⇒ $(x^2-\frac{9}{2}x)=\frac{3}{2}$

⇒ $(x^2-2(1)\frac{9}{4}x)=\frac{3}{2}$

⇒ $(x^2-2(1)\frac{9}{4}x) + (\frac{9}{4})^2=\frac{3}{2} + (\frac{9}{4})^2$

⇒ $(x-\frac{9}{4})^2=\frac{3}{2} + (\frac{81}{16})$

⇒ $(x-\frac{9}{4})^2=\frac{24}{16} + (\frac{81}{16})$

⇒ $(x-\frac{9}{4})^2=\frac{105}{16}$

⇒ $-2(x-\frac{9}{4})^2=-2(\frac{105}{16})$

⇒ $-2(x-\frac{9}{4})^2=-\frac{105}{8}$

⇒ $-2(x-\frac{9}{4})^2+\frac{105}{8} = 0$

Comparing this equation to $-2(x-p)^2 +q=0$ , $q =\frac{105}{8}$ .

Therefore ,Value of q is $\frac{105}{8}$ .