Question $n$ such that $n \geq 4$

Studdy Solution

STEP 1

Assumptions
1. The given equation is a quadratic equation in the form $av^2 + bv + c = 0$ .
2. We are looking for values of $n$ such that the equation has exactly one real solution.
3. A quadratic equation has exactly one real solution if and only if its discriminant is zero.
4. The discriminant of a quadratic equation $av^2 + bv + c = 0$ is given by $b^2 - 4ac$ .

STEP 2

Write down the discriminant of the given quadratic equation.
$Discriminant = b^2 - 4ac$

STEP 3

Identify the coefficients $a$ , $b$ , and $c$ from the given equation.
$a = n - 4, \quad b = 5, \quad c = -4$

STEP 4

Substitute the identified coefficients into the discriminant formula.
$Discriminant = 5^2 - 4(n - 4)(-4)$

STEP 5

Simplify the expression for the discriminant.
$Discriminant = 25 + 16(n - 4)$

STEP 6

Since we want the equation to have exactly one real solution, set the discriminant equal to zero.
$25 + 16(n - 4) = 0$

STEP 7

Solve for $n$ .
$16(n - 4) = -25$

STEP 8

Divide both sides of the equation by 16.
$n - 4 = \frac{-25}{16}$

STEP 9

Add 4 to both sides of the equation to isolate $n$ .
$n = \frac{-25}{16} + 4$

STEP 10

Convert 4 into a fraction with a denominator of 16 to combine with $\frac{-25}{16}$ .
$4 = \frac{64}{16}$

STEP 11

Add the fractions.
$n = \frac{-25}{16} + \frac{64}{16}$

STEP 12

Combine the numerators.
$n = \frac{-25 + 64}{16}$

STEP 13

Calculate the combined numerator.
$n = \frac{39}{16}$
The value of $n$ for which the equation has one real solution is $n = \frac{39}{16}$ .

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Parents Influencer program Contact Policy Terms