Question

Solve: x^2 - b^2 - = a(2x - a)

Grade 10, Quadratic Equations.

Answer 1

Answer:

$\Large\boxed{X=A+\sqrt{b^2+1}, X=A-\sqrt{b^2+1} }$

Step-by-step explanation:

Given:

x²-b²-=a(2x-a)

To solve this problem, first you have to isolate by the x or a from one side of the equation.

$\Large\boxed{\textnormal{SUBJECT: MATH}}$

$\Large\boxed{\textnormal{LESSON: SOLVE WITH QUADRATIC EQUATIONS}}$

Solutions:

First, used quadratic equations formula:

$\Large\boxed{\textnormal{QUADRATIC EQUATIONS FORMULA}}$

Expand by using with distributive property.

$\Large\boxed{\textnormal{DISTRIBUTIVE PROPERTY}}$

$\displaystyle A(B+C)=AB+AC$

$\displaystyle a(2x-a)$

A=A

B=2x

C=A

$\displaystyle a*2x-aa$

$\displaystyle 2xa-aa$

$\displaystyle A^2=AA$

$\displaystyle 2xa-a^2$

Rewrite the problem down.

$\displaystyle X^2-b^2-=2xa-a^2$

You have to add a² from both sides.

$\displaystyle x^2-b^2-+a^2=2xa-a^2+a^2$

Solve.

$\displaystyle x^2-b^2-+a^2=2xa$

Then, subtract 2xa from both sides.

$\displaystyle x^2-b^2-+a^2-2xa=2xa-2xa$

Solve. (Simplify/refine/solution.)

$\displaystyle x^2-2ax-b^2-1+a^2=0$

A=1

B=(-2a)

C=(-b²-1+a²)

$\displaystyle \frac{-(-2a)+\sqrt{(-2a)^2-4*1(-b^2-1+a^2)}}{2*1}$

Solve.

$\displaystyle \frac{-(-2a)+\sqrt{(-2a)^2-4*1(-b^2-1+a^2)}}{2*1}=\boxed{a+\sqrt{b^2+1}, \quad a-\sqrt{b^2+1} }$

$\Large\boxed{\boxed{a+\sqrt{b^2+1}, \quad a-\sqrt{b^2+1} }}$

Therefore, the correct answer is x=a+√b²+1, and x=a-√b²+1.

Answer 2

$\huge\boxed{SOLUTION}$

$x = a + \sqrt{b {}^{2} + 1 } \\ k = a - \sqrt{b {}^{2} + 1 }$

Further Explanation

According to the given question:-

$x {}^{2} - b {}^{2} - = (2x - a)$

Now,

$a(b + c) = ab + ac$

$a(2x - a)$

$(a = a)$

$(b = 2x)$

$(c = a)$

$(a°2x - aa)$

$(a {}^{2} = aa)$

$(2xa - a {}^{2} )$

Here. ...

$= x {}^{2} - b {}^{2} - = 2x {}^{2} a {}^{2} \\ = x {}^{2} - b {}^{2} - + a {}^{2} = 2xa - a {}^{2} + a {}^{2} \\ = x {}^{2} - b {}^{2} - + a {}^{2} = 2xa \\ = x {}^{2} - b {}^{2} - + a {}^{2} - 2xa = 2xa - 2xa$

Simplifying

$x {}^{2} - 2ax - b {}^{2} - 1 + a {}^{2} = 0 \\ = (a = 1) \\ = (b = ( - 2a)) \\ = (c) = ( - b {}^{2} - 1 + a { {}^{} }^{2}))$

Then,finally

$- ( - 2a) + \sqrt{( - 2a) {}^{2} - 4°1( - b {}^{2}1 + a {}^{2} } by \: 2°1$

$= a + \sqrt{b {}^{2} + 1} .. \sqrt{b {}^{2} + 1 }$