Question

Find all values of for which the quadratic equation has two real solutions.5x^2+9x+h=0Write your answer as an equality or inequality in terms of .

Answer 1

Given the equation:

$5x^2+9x+h=0$

we will find the value of h which the quadratic equation has two real solutions

We will use the discriminant of the equation (D)

$\begin{gathered} a=5,b=9,c=h \\ D=b^2-4ac \end{gathered}$

substitute with the values of a, b, and c

$\begin{gathered} D=9^2-4\cdot5\cdot h \\ D=81-20h \end{gathered}$

the quadratic equation has two real solutions when D≥0

So,

$\begin{gathered} 81-20h\ge0 \\ -20h\ge-81\rightarrow(\div-20) \\ \frac{-20h}{-20}\le\frac{-81}{-20} \\ \\ h\le4.05 \end{gathered}$

So, the answer will be:

$h\le4.05$