Question

For what value of k does the equation (2k+1)x^2+2x=10x-6 have two real and equal roots?

Answer 1

Answer:

$\displaystyle k = \frac{5}{6}$

Step-by-step explanation:

We are given the equation:

$\displaystyle (2k+1)x^2 + 2x = 10x - 6$

And we want to find the value of k such that the equation has two real and equivalent roots.

Since the equation is a quadartic, we can find its discriminant (symbolized by Δ). Recall that:

• If Δ < 0, we have no real roots (two complex roots).
• If Δ > 0, we have two real roots.
• And if Δ = 0, we have one real root, or two equivalent ones.

First, rewrite our equation:

$(2k+1)x^2 -8x + 6 =0$

The discriminant is given by:

$\displaystyle \Delta = b^2 -4ac$

In this case, b = -8, a = (2k + 1), and c = 6.

Therefore, the discriminant is given by:

$\displaystyle \Delta = (-8)^2 - 4(2k+1)(6)$

For it to have two equal roots, the discriminant must be zero. Hence:

$\displaystyle 0 = (-8)^2 - 4(2k+1)(6)$

Solve for k:

$\displaystyle \begin{aligned} \displaystyle 0 &= (-8)^2 - 4(2k+1)(6) \\ 0 &= 64 - 48k - 24 \\ 0 &= 40 - 48k \\ -40 &= -48k \\ \\ k &= \frac{5}{6} \end{aligned}$

Hence, the value of k is 5/6.