Answer:
c / 2x²=32
Step-by-step explanation:
so we have 2x²=32
x²=32/2
x²=16
x=√16 and x= -√16
x=4 and x= -4
Answer:
11j^2
Step-by-step explanation:
both are divisible by 11 and j^2
I think commutative property but if not then distributive property
Answer:
![x_1=1+\sqrt{89}\\\\x_2=1-\sqrt{89}](https://tex.z-dn.net/?f=x_1%3D1%2B%5Csqrt%7B89%7D%5C%5C%5C%5Cx_2%3D1-%5Csqrt%7B89%7D)
Step-by-step explanation:
Apply Distributive property:
![(x+7)(x-9)=25\\\\x^2-9x+7x-63=25](https://tex.z-dn.net/?f=%28x%2B7%29%28x-9%29%3D25%5C%5C%5C%5Cx%5E2-9x%2B7x-63%3D25)
Add like terms and then add 63 to both sides of the equation:
![x^2-2x-63=25\\\\x^2-2x-63+63=25+63\\\\x^2-2x=88](https://tex.z-dn.net/?f=x%5E2-2x-63%3D25%5C%5C%5C%5Cx%5E2-2x-63%2B63%3D25%2B63%5C%5C%5C%5Cx%5E2-2x%3D88)
Pick the coefficient of the x term, divide it by 2 and square it:
![(\frac{2}{2})^2=1](https://tex.z-dn.net/?f=%28%5Cfrac%7B2%7D%7B2%7D%29%5E2%3D1)
Add it to both sides of the equation:
![x^2-2x+1=88+1](https://tex.z-dn.net/?f=x%5E2-2x%2B1%3D88%2B1)
Rewriting the left side as a squared binomial, we get:
![(x-1)^2=89](https://tex.z-dn.net/?f=%28x-1%29%5E2%3D89)
Apply square root to both sides:
![\sqrt{(x-1)^2}=\±\sqrt{89}\\\\x-1=\±\sqrt{89}](https://tex.z-dn.net/?f=%5Csqrt%7B%28x-1%29%5E2%7D%3D%5C%C2%B1%5Csqrt%7B89%7D%5C%5C%5C%5Cx-1%3D%5C%C2%B1%5Csqrt%7B89%7D)
And finally we need to add 1 to both sides of the equation. Then we get:
![x-1+1=\±\sqrt{89}+1\\\\x=\±\sqrt{89}+1\\\\\\x_1=1+\sqrt{89}\\\\x_2=1-\sqrt{89}](https://tex.z-dn.net/?f=x-1%2B1%3D%5C%C2%B1%5Csqrt%7B89%7D%2B1%5C%5C%5C%5Cx%3D%5C%C2%B1%5Csqrt%7B89%7D%2B1%5C%5C%5C%5C%5C%5Cx_1%3D1%2B%5Csqrt%7B89%7D%5C%5C%5C%5Cx_2%3D1-%5Csqrt%7B89%7D)