Question

abx^2=(a-b)^2(x+1)
1) find x in terms of a and b
2)find a in terms of x and b

Answer 1

Step-by-step explanation:

We have,

$ab {x}^{2} = (a - b)^{2} (x + 1)$

$\implies \: (ab) {x}^{2} = (a - b)^{2}x + (a - b)^{2}$

$\implies \: (ab) {x}^{2} - (a - b)^{2}x - (a - b)^{2} = 0$

Applying quadratic formula,

$\implies \: x = \frac{ - \{ - (a - b)^{2} \} \pm \sqrt{ \{ - (a - b)^{2}\} ^{2} - 4.(ab). \{ - (a - b)^{2}\}} }{2.(ab)}$

$\implies \: x = \frac{ (a - b)^{2} \pm \sqrt{ (a - b)^{4} + 4.(ab). (a - b)^{2}} }{2.(ab)}$

$\implies \: x = \frac{ (a - b)^{2} \pm \sqrt{ (a - b)^{2}\{ (a - b)^{2} + 4ab \}} }{2.(ab)}$

$\implies \: x = \frac{ (a - b)^{2} \pm (a - b) \sqrt{ (a - b)^{2} + 4ab } }{2ab}$

$\implies \: x = \frac{ (a - b)^{2} \pm (a - b) \sqrt{ a^{2} + b^{2} - 2ab + 4ab } }{2ab}$

$\implies \: x = \frac{ (a - b)^{2} \pm (a - b) \sqrt{ a^{2} + b^{2} + 2ab } }{2ab}$

$\implies \: x = \frac{ (a - b)^{2} \pm (a - b) \sqrt{ (a + b)^{2} } }{2ab}$

$\implies \: x = \frac{ (a - b)^{2} \pm (a - b) (a + b) }{2ab}$

$\implies \: x = \frac{ (a - b)^{2} \pm (a ^{2} - b^{2} ) }{2ab}$

Either $x = \frac{ (a - b)^{2} + (a ^{2} - b^{2} ) }{2ab}$ or $x = \frac{ (a - b)^{2} - (a ^{2} - b^{2} ) }{2ab} \\ [\tex]</p><p>[tex] \implies x = \frac{ a^{2} + b^{2} - 2ab + a ^{2} - b^{2} }{2ab} \: \: or \: \: x = \frac{ a^{2} + b^{2} - 2ab - a ^{2} + b^{2} }{2ab}$

$\implies x = \frac{ a^{2} - 2ab + a ^{2} }{2ab} \: \: or \: \: x = \frac{ b^{2} - 2ab + b^{2} }{2ab}$

$\implies x = \frac{ 2a^{2} - 2ab }{2ab} \: \: or \: \: x = \frac{ 2 b^{2} - 2ab }{2ab}$

$\implies x = \frac{ 2a(a -b ) }{2ab} \: \: or \: \: x = \frac{ 2 b(b - a ) }{2ab}$

$\implies x = \frac{a -b }{b} \: \: or \: \: x = \frac{ b - a }{a}$

Now, from the above values of x, we get,

$\implies x = \frac{a }{b} - 1 \: \: or \: \: x = \frac{ b }{a} - 1$

$\implies x + 1 = \frac{a }{b} \: \: or \: \: x + 1= \frac{ b }{a}$

$\implies a = b( x + 1) \: \: or \: \: a = \frac{ b }{x + 1}$