Question

#11x^2+2x=1a) find the value of b^2- 4ac, and b) determine how many real solutionsquadratic has. Do not solve the quadratic equation.

Answer 1

Step 1

Given;

$x^2+2x=1$

Step 2

A) Find the value of b²-4ac

$\begin{gathered} From\text{ the equation} \\ a=1 \\ b=2 \\ c=-1 \end{gathered}$$\begin{gathered} 2^2-4(1)(-1)=4+4=8 \\ b^2-4ac=8 \end{gathered}$

Step 3

B) Determine how many real solutions quadratic has.

$\begin{gathered} Since\text{ D is greater than is equal to 8 which is greater than 0.} \\ The\text{ quadratic equation has two distinct real roots.} \end{gathered}$