For what value of c does the quadratic equation 0.5x^2 - 3x + c = 0 have two real solution… — ACT

For what value of $c$ does the quadratic equation $0.5x^2 - 3x + c = 0$ have two real solutions that differ by exactly 4?

2.5

The correct value is 2.5. By utilizing the formula for the difference of the roots,

\frac{\sqrt{b^2 - 4ac}}{|a|} = 4

, and substituting

a = 0.5

and

b = -3

, we get

\frac{\sqrt{9 - 2c}}{0.5} = 4

. This simplifies to

\sqrt{9 - 2c} = 2

, which squares to

9 - 2c = 4

. Solving for

c

yields 2.5.

Identify the coefficients and apply the relationship for the difference between two roots.

For a quadratic equation

ax^2 + bx + c = 0

, the roots

x_1

and

x_2

satisfy

|x_1 - x_2| = \frac{\sqrt{b^2 - 4ac}}{|a|}

. Here,

a = 0.5

b = -3

, and the difference is 4.

This formula relates the difference of the roots directly to the coefficients of the quadratic equation.

Substitute the given values into the formula and solve for

c

Substituting the values gives

\frac{\sqrt{(-3)^2 - 4(0.5)c}}{|0.5|} = 4 \Rightarrow \frac{\sqrt{9 - 2c}}{0.5} = 4

. Multiplying both sides by 0.5 yields

\sqrt{9 - 2c} = 2

. Squaring both sides gives

9 - 2c = 4

Simplifying the equation isolates the variable

c

under the radical.

Complete the algebraic isolation to find the final value of

c

2c = 5 \Rightarrow c = 2.5

This final step solves the linear equation for

c

Using the discriminant and properties of roots to solve quadratic equations with given constraints.

Question