Chapter ISG, Problem 9.7.15P

To determine

To find value of k.

Answer to Problem 9.7.15P

  $k=-\frac{1}{4}$ is value at which system of equation having exact one solution

Explanation of Solution

Given information:

  $y=x^2+2$

  $y=3x+k$

Calculation:

  $y=x^2+\mathrm{2..............}(1)$

  $y=3x+k\mathrm{..........}(2)$

Putting value of y from equation (2) in (1)

  $3x+k=x^2+2$

On solving,

  $0=x^2+2-3x-k$

On solving,

  $0=x^2-3x+2-k$

Here,

  $a=1$

  $b=-3$

  $b=2-k$

For exact one solution

  $b^2-4ac=0$

Put value of a, b and c

  ${(-3)}^2-4(1)(2-k)=0$

On solving,

  $9-4(2-k)=0$

On solving,

  $4(2-k)=9$

On solving,

  $(2-k)=\frac{9}{4}$

On solving,

  $-k=\frac{9}{4}-2$

On solving,

  $-k=\frac{1}{4}$

On solving,

  $k=-\frac{1}{4}$

Hence, $k=-\frac{1}{4}$ is value at which system of equation having exact one solution