Multiple choice

(a) Show that if x is real, the expression $\frac { { x }^{ 2 }-bc }{ 2x-b-c } $ has no real values between b and c. (b) For real x, the function $\frac { (x-a)(x-b) }{ x-c } $ will assume all real values provided (i) $a>b>c$ (ii) $ac>b$ (iv) $a

  1.  Positive

  2. Negative

  3. Neal and of opposite sign

  4. Imaginary
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If alpha and beta are the roots of the first quadratic, then alpha beta < 0 because the roots have opposite signs. Expanding the second equation shows that its root product is also alpha beta, which is negative, while its discriminant is positive. Therefore, its roots are real and of opposite signs.

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