Question

Hey, (x+2)^2 =(x-5)^2+7,how do I solve for x? At first here we have to use the formulas (a+b)^2 and (a-b)^2 for solving (x+2)^2 and (x-5)^2 …we know (a+b)^2= a^2+2ab+b^2 and (a-b)^2= a^2–2ab+b^2 So using these formulas solve for (x+2)^2 and (x-5)^2 (x+2)^2= x^2+4x+4 (x-5)^2=x^2–10x+25 So solve the equation (x+2)^2=(x-5)^2+7 =>x^2+4x+4=x^2–10x+25+7 =>4x+4=32–10x =>4x+10x=32–4 =>14x=28 =>x=28/14 =>x=2 If you want to check if your answer is correct or not then you can put the value of x in the equation. So putting the value of x in the equation we get (2+2)^2=(2–5)^2+7 =>4^2=(—3)^2+7 =>16=16 Thus,our value of x is correct So the value of x =2 I hope you may get some help.. Thank you

Answer 1

Explanation:

Hey,

(x+2)^2 =(x-5)^2+7,how do I solve for x?

At first here we have to use the formulas (a+b)^2 and (a-b)^2 for solving (x+2)^2 and (x-5)^2 …we know (a+b)^2= a^2+2ab+b^2 and

(a-b)^2= a^2–2ab+b^2

So using these formulas solve for (x+2)^2 and (x-5)^2

(x+2)^2= x^2+4x+4

(x-5)^2=x^2–10x+25

So solve the equation

(x+2)^2=(x-5)^2+7

=>x^2+4x+4=x^2–10x+25+7

=>4x+4=32–10x

=>4x+10x=32–4

=>14x=28

=>x=28/14

=>x=2

If you want to check if your answer is correct or not then you can put the value of x in the equation.

So putting the value of x in the equation we get

(2+2)^2=(2–5)^2+7

=>4^2=(—3)^2+7

=>16=16

Thus,our value of x is correct

So the value of x =2

I hope you may get some help..

Thank you

Answer 2

Answer:

Explanation:The quadratic formula helps you solve quadratic equations, and is probably one of the top five formulas in math.  We’re not big fans of you memorizing formulas, but this one is useful (and we think you should learn how to derive it as well as use it, but that’s for the second video!).

If you have a general quadratic equation like this:

ax^2+bx+c=0ax  

2

+bx+c=0a, x, squared, plus, b, x, plus, c, equals, 0

Then the formula will help you find the roots of a quadratic equation, i.e. the values of xxx where this equation is solved.

The quadratic formula

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}x=  

2a

−b±  

b  

2

−4ac

​  

 

​  

x, equals, start fraction, minus, b, plus minus, square root of, b, squared, minus, 4, a, c, end square root, divided by, 2, a, end fraction

It may look a little scary, but you’ll get used to it quickly!

Practice using the formula now.

Worked example

First we need to identify the values for a, b, and c (the coefficients). First step, make sure the equation is in the format from above, ax^2 + bx + c = 0ax  

2

+bx+c=0a, x, squared, plus, b, x, plus, c, equals, 0:

x^2+4x-21=0x  

2

+4x−21=0x,