Question

For which values of a will f(x)=0 .....have non real values

Answer 1

Step 1:

Write the function

`f(x)=-2x^2\text{ + 4ax + 3}`

Step 2:

From a quadratic formula

A negative discriminant indicates that neither of the solutions are real numbers.

`\begin{gathered} \text{Discriminant D = b}^2\text{ - 4ac} \\ \end{gathered}`

Step 3

for non real values , D < 0

`b^2\text{ - 4ac < 0}`

Step 4:

From the general equation of the quadratic equation

`\begin{gathered} ax^2\text{ + bx + c = 0} \\ a\text{ = -2} \\ b\text{ = 4a} \\ c\text{ = 3} \end{gathered}`

Step 5:

`\begin{gathered} b^2\text{ - 4ac < 0} \\ (4a)^2\text{ - 4(-2)(3) < 0} \\ 16a^2\text{ + 24 < 0} \\ 16a^2\text{ < -}24 \\ a^2\text{ < }(-24)/(16) \\ a^2\text{ < }(-3)/(2) \\ a\text{ <}\pm\text{ }\sqrt[]{(-3)/(2)} \\ a\text{ < i}\sqrt[]{(3)/(2)\text{ }}\text{ or a < -}\sqrt[]{(3)/(2)} \end{gathered}`

Final answer

False