Question

The function Q(x)=2x^2-kx+18. For what value of k does Q(x) have one distinct real solution? (NEEDS TO BE DONE WITHIN 2 HOURS!)

Answer 1

Answer:

k = 12

Step-by-step explanation:

Given:

The equation $Q(x)=2x^2-kx+18$

To find:

Value of $k = ?$ for which the given equation has one distinct real solution.

Solution:

The given equation is a quadratic equation.

There are always two solutions of a quadratic equation.

For the equation: $ax^{2} +bx+c=0$ to have one distinct solution:

$b^2 - 4ac = 0$

Here,

a = 2,

b = -k and

c = 18

Putting the values, we get:

$(-k)^2 - 4\times 2\times 18 = 0\\\Rightarrow k^2 = 18\times 8\\\Rightarrow k^2 =144\\\Rightarrow k = 12$

The equation becomes:

$Q(x)=2x^2-12x+18$

And the one root is:

$2(x^2-6x+9 ) = 0\\\Rightarrow 2(x-3)^2=0\\\Rightarrow x = 3$