Condition for Equal Roots

Given,

$$ax^2-2bx+15=0$$

has repeated root $$\alpha$$.

Hence, discriminant is zero:

$$(-2b)^2-4(a)(15)=0$$

$$4b^2-60a=0$$

$$b^2=15a$$

Now, for repeated root,

$$\alpha=\frac{2b}{2a}=\frac ba$$

Using

$$b^2=15a$$

we get

$$\alpha=\frac{15}{b}$$

Also, $$\alpha$$ and $$\beta$$ are roots of

$$x^2-2bx+21=0$$

Hence,

$$\alpha+\beta=2b,\qquad \alpha\beta=21$$

Since $$\alpha$$ is a root,

$$\alpha^2-2b\alpha+21=0$$

Substituting

$$\alpha=\frac{15}{b}$$

we get

$$\left(\frac{15}{b}\right)^2-2b\left(\frac{15}{b}\right)+21=0$$

$$\frac{225}{b^2}-30+21=0$$

$$\frac{225}{b^2}=9$$

$$b^2=25$$

Now,

$$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$$

$$=(2b)^2-2(21)$$

$$=4(25)-42$$

$$=58$$

Hence, $$\boxed{58}$$.

We are given two real numbers $$a,\,b$$ with $$a\neq 0$$ such that the quadratic equation $$ax^{2}-2bx+5=0$$ possesses a repeated (double) root $$\alpha$$. For a quadratic $$Ax^{2}+Bx+C=0$$ to have a repeated root, its discriminant must vanish. The discriminant formula is $$\Delta=B^{2}-4AC$$, so we write

$$\Delta = (-2b)^{2}-4\,(a)\,(5)=0.$$

Evaluating each term, we get

$$4b^{2}-20a=0 \;\Longrightarrow\; b^{2}-5a = 0 \;\Longrightarrow\; b^{2}=5a. \quad -(1)$$

The value of the repeated root of a quadratic $$Ax^{2}+Bx+C=0$$ is $$-\dfrac{B}{2A}$$. In our first equation the coefficients are $$A=a,\;B=-2b$$, hence

$$\alpha = -\dfrac{-2b}{2a}=\dfrac{2b}{2a}=\dfrac{b}{a}. \quad -(2)$$

This same number $$\alpha$$ is also a root of the second quadratic equation

$$x^{2}-2bx-10=0.$$

Therefore $$\alpha$$ satisfies

$$\alpha^{2}-2b\alpha-10=0. \quad -(3)$$

Substituting $$\alpha=\dfrac{b}{a}$$ from (2) into (3) gives

$$\left(\dfrac{b}{a}\right)^{2}-2b\left(\dfrac{b}{a}\right)-10=0.$$

Simplify each term:

$$\dfrac{b^{2}}{a^{2}}-\dfrac{2b^{2}}{a}-10=0.$$

To clear denominators we multiply every term by $$a^{2}$$:

$$b^{2}-2b^{2}a-10a^{2}=0. \quad -(4)$$

We now use the earlier relation (1): $$b^{2}=5a$$. Substituting this into (4) gives

$$5a-2(5a)a-10a^{2}=0 \;\Longrightarrow\; 5a-10a^{2}-10a^{2}=0.$$

Combining like terms,

$$5a-20a^{2}=0 \;\Longrightarrow\; 5a\,(1-4a)=0.$$

Because $$a\neq 0$$, we must have

$$1-4a=0 \;\Longrightarrow\; a=\dfrac14. \quad -(5)$$

Using (1) again, we find

$$b^{2}=5a=5\!\left(\dfrac14\right)=\dfrac54.$$

Hence $$b=\pm\dfrac{\sqrt5}{2}.$$ (Either sign will yield the same final value we seek.) Now, from (2),

$$\alpha=\dfrac{b}{a}=b\div\dfrac14=4b,$$

so

$$\alpha^{2}=(4b)^{2}=16b^{2}=16\left(\dfrac54\right)=20. \quad -(6)$$

Let $$\beta$$ be the other root of the second equation $$x^{2}-2bx-10=0$$. For a quadratic $$x^{2}+px+q=0$$, the sum and product of roots are $$-p$$ and $$q$$ respectively. Here the coefficient of $$x$$ is $$-2b$$ and the constant term is $$-10$$, so

$$\alpha+\beta=2b,\qquad \alpha\beta=-10. \quad -(7)$$

We want $$\alpha^{2}+\beta^{2}$$. We use the algebraic identity

$$\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta. \quad -(8)$$

Substituting from (7) into (8),

$$\alpha^{2}+\beta^{2}=(2b)^{2}-2(-10)=4b^{2}+20.$$

But $$b^{2}=\dfrac54$$, so

$$4b^{2}+20=4\left(\dfrac54\right)+20=5+20=25. \quad -(9)$$

Thus,

$$\alpha^{2}+\beta^{2}=25.$$

Hence, the correct answer is Option A.