Question

If $a\ne b$ the roots of the equation $(x-a)(x-b)=b^2$ are

• A. real and distinct
• B. real and equal
• C. real
• D. imaginary

Answer 1

Answer: (A)

real and distinct

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

$x^2-(a+b)x+ab-b^2=0$

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

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

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

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

Since

$(a-b{)}^2\ge 0$ and $b^2\ge 0$.

Hence

$(a-b{)}^2+4b^2>0$.

Therefore,

$B^2-4AC>0$

Hence $D>0$.

Thus the roots are real and distinct.