[SOLVED] Find the value of constant k in tangent

**Question:**

Find the value of the constant $\displaystyle k$ for which the line $\displaystyle y + 2x = k$ is a tangent to the curve $\displaystyle y = x^2 - 6x + 14.$

**Attempt:**

$\displaystyle y = x^2 - 6x + 14$

$\displaystyle y + 2x = k$

$\displaystyle y = k - 2x$

$\displaystyle = x^2 - 6x + 2x + 14 - k$

$\displaystyle = x^2 - 4x + 14 - k$

$\displaystyle a = 1$, $\displaystyle b = -4$, $\displaystyle c = 14$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$\displaystyle x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4\times 1\times 14}}{2\times 1}$

I can't continue because Im getting a negative value inside the square root, where did I go wrong?

Re: [SOLVED] Find the value of constant k in tangent

Quote:

Originally Posted by

**looi76** **Question:**

Find the value of the constant $\displaystyle k$ for which the line $\displaystyle y + 2x = k$ is a tangent to the curve $\displaystyle y = x^2 - 6x + 14.$

**Attempt:**

$\displaystyle y = x^2 - 6x + 14$

$\displaystyle y + 2x = k$

$\displaystyle y = k - 2x$

$\displaystyle = x^2 - 6x + 2x + 14 - k$

$\displaystyle = x^2 - 4x + 14 - k$

$\displaystyle a = 1$, $\displaystyle b = -4$, $\displaystyle c = 14$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$\displaystyle x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4\times 1\times 14}}{2\times 1}$

I can't continue because Im getting a negative value inside the square root, where did I go wrong?

Here is how you do it.. Easy and Simple

Line: **y = k-2x**

Curve: **y = x²-6x+14**

__Workings__

**k-2x = x²-6x+14**

x²-6x+2x+14-k = 0

x²-4x+14-k = 0

a=1, b=-4, c=(14-k)

[I]Use this Equation[/I] ---- **b²-4ac**

**b²-4ac**

Replace..

=(-4)²-[4(14-k)]

=16-(56-4k)

=16-56+4k

4k = -16+56

4k = 40

k = 40/4

k = 10 <---- Here's your Answer!!!