Question

x2-2x-(r2-1) =0 find roots
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Answer 1

$\large\underline\blue{\bold{Given\:Question - }}$

• Solve - x² - 2x - (r² - 1) = 0

$\large\underline\purple{\bold{Solution :- }}$

$\longmapsto\tt \: {x}^{2} - 2x - ( {r}^{2} - 1) = 0$

$\longmapsto\tt \: {x}^{2} - 2x - (r + 1)(r - 1) = 0$

$\longmapsto\tt \: {x}^{2} - (1 + 1)x - (r + 1)(r - 1) = 0$

$\longmapsto\sf\: {x}^{2} - (1 + 1 + r - r)x - (r + 1)(r - 1) = 0$

$\tt\: {x}^{2} - \bigg((r + 1) - (r - 1) \bigg)x - (r + 1)(r - 1) = 0$

$\tt \: {x}^{2} - (r + 1)x + (r - 1)x - (r + 1)(r - 1) = 0$

$\longmapsto\tt \: x(x - r - 1) + (r - 1)(x - r - 1) = 0$

$\longmapsto\tt \: (x - r - 1)(x + r - 1) = 0$

$\bf\implies \: \red{x \: = r + 1} \: \: or \: \: \red{x \: = 1 - r}$

Additional Information :-

• Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots depends upon Discriminant.

• If Discriminant, D > 0, equation has real and unequal roots.

• If Discriminant, D < 0, equation has no real roots or imaginary roots or complex roots.

• If Discriminant, D = 0, equation has real and equal roots.

• If Discriminant, D is positive and perfect square, equation has real and unequal rational roots.

• If Discriminant, D is positive and not perfect square, equation has real and unequal irrational roots.