Chapter A.6, Problem 146AYU

To determine

To find: The three quadratic equations: one having two distinct solutions, one having no real solution, and one having exactly one real solution.

Answer to Problem 146AYU

$3x^2+5x-8=0$

$a{x}^2 + bx + c = 0$ Compare to standard form.

With $a=3$, $b=5$, and $c=-8$, the discriminant is $b^2-4ac={\left(5\right)}^2-4\left(3\right)\left(-8\right)=121$.

Since $b^2-4ac>0$, there are two unequal real solutions.

$9x^2-30x+25=0$

$a{x}^2 + bx + c = 0$ Compare to standard form.

With $a=9$, $b=-30$, and $c=25$, the discriminant is $b^2-4ac={\left(-30\right)}^2-4\left(9\right)\left(25\right)=0$.

Since $b^2-4ac=0$, there is repeated real solution.

$x^2+5x+7=0$

$a{x}^2 + bx + c = 0$ Compare to standard form.

With $a=1, b=5$, and $c=7$, the discriminant is $b^2-4ac={\left(5\right)}^2-4\left(1\right)\left(7\right)=-3$.

Since $b^2-4ac<0$, there is no real solution.

Explanation of Solution

$3x^2+5x-8=0$

$a{x}^2 + bx + c = 0$ Compare to standard form.

With $a=3$, $b=5$, and $c=-8$, the discriminant is $b^2-4ac={\left(5\right)}^2-4\left(3\right)\left(-8\right)=121$.

Since $b^2-4ac>0$, there are two unequal real solutions.

$9x^2-30x+25=0$

$a{x}^2 + bx + c = 0$ Compare to standard form.

With $a=9$, $b=-30$, and $c=25$, the discriminant is $b^2-4ac={\left(-30\right)}^2-4\left(9\right)\left(25\right)=0$.

Since $b^2-4ac=0$, there is repeated real solution.

$x^2+5x+7=0$

$a{x}^2 + bx + c = 0$ Compare to standard form.

With $a=1, b=5$, and $c=7$, the discriminant is $b^2-4ac={\left(5\right)}^2-4\left(1\right)\left(7\right)=-3$.

Since $b^2-4ac<0$, there is no real solution.