b Find the values of $$ \alpha $$ for which the quadratic equation $$ 4 x^{2}-(\alpha-12) x+(\alpha-7)=0 $$ has equal roots

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We start with the quadratic equation

$$
4x^2 - (\alpha - 12)x + (\alpha - 7) = 0.
$$

For a quadratic equation $$ax^2 + bx + c = 0$$ to have equal (repeated) roots, the discriminant must be zero. The discriminant is given by

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

Here, the coefficients are

$$
a = 4,\quad b = -(\alpha - 12) = 12 - \alpha,\quad c = \alpha - 7.
$$

Setting the discriminant to zero:

$$
(12 - \alpha)^2 - 4 \cdot 4 \cdot (\alpha - 7) = 0.
$$

Expanding the square and the product:

$$
(12 - \alpha)^2 = \alpha^2 - 24\alpha + 144,
$$
$$
4 \cdot 4 \cdot (\alpha - 7) = 16(\alpha - 7).
$$

Thus, the equation becomes

$$
\alpha^2 - 24\alpha + 144 - 16(\alpha - 7) = 0.
$$

Next, distribute and simplify:

$$
\alpha^2 - 24\alpha + 144 - 16\alpha + 112 = 0,
$$
$$
\alpha^2 - 40\alpha + 256 = 0.
$$

This is a quadratic equation in $$\alpha$$. We can solve it using the quadratic formula:

$$
\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
$$

with $$a = 1$$, $$b = -40$$, and $$c = 256$$. First, compute the discriminant:

$$
(-40)^2 - 4(1)(256) = 1600 - 1024 = 576.
$$

Taking the square root:

$$
\sqrt{576} = 24.
$$

Now, solve for $$\alpha$$:

$$
\alpha = \frac{40 \pm 24}{2}.
$$

This gives two solutions:

$$
\alpha = \frac{40 + 24}{2} = \frac{64}{2} = 32,
$$
$$
\alpha = \frac{40 - 24}{2} = \frac{16}{2} = 8.
$$

Thus, the values of $$\alpha$$ for which the quadratic equation has equal roots are

$$
\boxed{8 \text{ and } 32}.
$$
The values of $$\alpha$$ that make the quadratic equation $$4x^{2}-(\alpha-12)x+(\alpha-7)=0$$ have equal roots are 8 and 32.

b Find the values of for which the quadratic equation has equal roots

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