Question

solve by using formula method
x^2-2ax+(a+1)(a-1)=0​

Answer 1

$\bold \red \star \: \large \underline{Given:-} \begin{cases} \sf{ \underline{Our \: \: equation :-}} \\ \sf{ {x}^{2} - 2ax + (a + 1)(a - 1) = 0.} \end{cases}$

$\\ \\ \bold \red \star \: \large \underline{Solution:-}$

${x}^{2} - 2ax + (a + 1)(a - 1) = 0. \\ \\ or \: \: \: \: \: {x}^{2} - 2ax + ( {a}^{2} - {1}^{2} ) = 0.$ \\ \\

Let's Comparing with General Equation:-

$a {x}^{2} + bx + c = 0.$ $\\ where \: as, \: \\ \\ \: \: \: \: \: \: \: \: \: a = 1. \: \: \: \: \: b = - 2a .\: \: c = ( {a}^{2} - 1).$

$\rightarrow D= {b}^{2} - 4ac .\\ \\ \: \: \: \: \: \: \: \: \: \: \: ={ ( - 2a) }^{2} - 4(1)( {a}^{2} - 1). \\ \\ \: \: \: \: \: \: \: \: \: \: \: = \cancel{ 4 {a}^{2}} - \cancel {4 {a}^{2} } + 4. \\ \\ \: \: \: \: \: \: \: \: \: \: \: = 4.$

Now,

Let's Find out The value of x.

$x = \frac{ - b ± \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ \: \: \: \: = \frac{2a ± \sqrt{4} }{2} \\ \\ \: \: \: \: = \frac{2a±2 }{2} \\ \\ \: \: \: \: = \frac{ \cancel{2}(a±1) }{ \cancel2} \\ \\ \: \: \: \: = a±1.$

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So,

We have two value of a.

$a + 1 = 0. \: \: \: \: \: \: \: \: or \: \: \: \: \: \: \: \: a - 1 = 0.\\ \\ a = - 1. \: \: \: \: \: \: \: \: \: \: \: or \: \: \: \: \: \: \: \: \: a = 1.$