Question

Solve the following equations:
$\sqrt{x^2+ax-1}-\sqrt{x^2+bx-1}=\sqrt{a}-\sqrt{b}$.

• A. $-2,\frac{(\sqrt{a}-\sqrt{b}{)}^2+3}{(\sqrt{a}+\sqrt{b}{)}^2-3}$
• B. $-1,\frac{(\sqrt{a}-\sqrt{b}{)}^2-4}{(\sqrt{a}+\sqrt{b}{)}^2+4}$
  
• C. $\frac{1}{2},\frac{(\sqrt{a}-\sqrt{b}{)}^2+4}{(\sqrt{a}+\sqrt{b}{)}^2-4}$
• D. $1,\frac{(\sqrt{a}-\sqrt{b}{)}^2+4}{(\sqrt{a}+\sqrt{b}{)}^2-4}$

Answer 1

Answer: (D)

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

$\sqrt{x^2+ax-1}-\sqrt{x^2+bx-1}=\sqrt{a}-\sqrt{b}$ ........(i)

$\sqrt{x^2+ax-1}-\sqrt{x^2+bx-1}\times \frac{\sqrt{x^2+ax-1}+\sqrt{x^2+bx-1}}{\sqrt{x^2+ax-1}+\sqrt{x^2+bx-1}}=\sqrt{a}-\sqrt{b}\times \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}\frac{x^2+ax-1-x^2-bx+1}{\sqrt{x^2+ax-1}+\sqrt{x^2+bx-1}}=\frac{a-b}{\sqrt{a}+\sqrt{b}}\frac{(a-b)x}{\sqrt{x^2+ax-1}+\sqrt{x^2+bx-1}}=\frac{a-b}{\sqrt{a}+\sqrt{b}}(\sqrt{a}+\sqrt{b})x=\sqrt{x^2+ax-1}+\sqrt{x^2+bx-1}$

Squaring both sides, we get

$(a+b+2\sqrt{a}\sqrt{b})x^2=x^2+ax-1+x^2+bx-1+2\sqrt{x^2+ax-1}\sqrt{x^2+bx-1}$ ......(ii)

Squaring (i)

$a+b-2\sqrt{a}\sqrt{b}=x^2+ax-1+x^2+bx-1-2\sqrt{x^2+ax-1}\sqrt{x^2+bx-1}$ .........(iii)

Adding (ii) and (iii)

$(a+b+2\sqrt{a}\sqrt{b})x^2+a+b-2\sqrt{a}\sqrt{b}=4x^2+2(a+b)x-4$

$\Rightarrow (4-{(\sqrt{a}+\sqrt{b})}^2)x^2+2(a+b)x-(4+{(\sqrt{a}-\sqrt{b})}^2)=0$ ......(iv)

But $(4+{(\sqrt{a}-\sqrt{b})}^2)-(4-{(\sqrt{a}+\sqrt{b})}^2)=2(a+b)$

Substituting in (iv), we have

$(4-{(\sqrt{a}+\sqrt{b})}^2)x^2+(4+{(\sqrt{a}-\sqrt{b})}^2)x-(4-{(\sqrt{a}+\sqrt{b})}^2)x-(4+{(\sqrt{a}-\sqrt{b})}^2)=0\Rightarrow (4-{(\sqrt{a}+\sqrt{b})}^2)x(x-1)+(4+{(\sqrt{a}-\sqrt{b})}^2)(x-1)=0\Rightarrow x=1,\frac{{(\sqrt{a}-\sqrt{b})}^2+4}{{(\sqrt{a}+\sqrt{b})}^2-4}$