## Intermediate Algebra (12th Edition)

$b=\{ -10,10 \}$
$\bf{\text{Solution Outline:}}$ To find the value of $b$ such that the given equation, $p^2+bp+25=0 ,$ will have $1$ rational solution, equate the discriminant to $0$ and solve for the variable. $\bf{\text{Solution Details:}}$ In the equation above, $a= 1 ,$ $b= b ,$ and $c= 25 .$ Using the Discriminant Formula which is given by $b^2-4ac,$ the discriminant is \begin{array}{l}\require{cancel} (b)^2-4(1)(25) \\\\= b^2-100 .\end{array} Equating the discriminant to $0$ so that the given equation will have $1$ rational solution, then \begin{array}{l}\require{cancel} b^2-100=0 \\\\ b^2=100 .\end{array} Taking the square root of both sides, then \begin{array}{l}\require{cancel} b=\pm\sqrt{100} \\\\ b=\pm10 .\end{array} Hence, the given equation has $1$ rational solution when $b=\{ -10,10 \} .$