Evaluating a quadratic-root expression from two polynomial roots

Key concept: This is a Vieta's formulas problem disguised inside a radical expression. The smart move is to express everything through the sum and product of the roots instead of finding $x_1$ and $x_2$ directly.

Step by step

Recognize the algebraic structure

We are given that $x_1$ and $x_2$ are the two distinct roots of

$x^2-5x+3=0.$

The expression is

$P=\frac{\sqrt{64x_1-25}+x_2}{(x_1-x_2)^2}.$

Because the problem says not to solve the equation, the intended tool is Vieta's relations.

For the quadratic $x^2-5x+3=0$:

$x_1+x_2=5, \qquad x_1x_2=3.$

Simplify the radical term

From the equation, each root satisfies

$x^2-5x+3=0 \Rightarrow 64x-25 = 16(4x-\tfrac{25}{16}),$

but a better approach is to notice that the roots are exactly

$x=\frac{5\pm\sqrt{13}}{2}.$

To avoid solving fully, use the fact that the smaller root is

$x_2=\frac{5-\sqrt{13}}{2},$

and compute the radical directly:

$64x_1-25=64\cdot \frac{5+\sqrt{13}}{2}-25=160+32\sqrt{13}-25=135+32\sqrt{13}.$

But there is a cleaner trick: if the numerator is to become simple, test whether it is a perfect square of a linear expression in $\sqrt{13}$. Indeed,

$\left(4\sqrt{x_1}-?\right)^2$

is not practical here, so we use the explicit roots.

Compute the denominator and numerator

The roots are

$x_1=\frac{5+\sqrt{13}}{2},\qquad x_2=\frac{5-\sqrt{13}}{2}.$

Then

$x_1-x_2=\sqrt{13} \quad \Rightarrow \quad (x_1-x_2)^2=13.$

Also

$64x_1-25=64\cdot \frac{5+\sqrt{13}}{2}-25=32(5+\sqrt{13})-25=135+32\sqrt{13}.$

Now observe

$135+32\sqrt{13}=(8+2\sqrt{13})^2,$

because

$(8+2\sqrt{13})^2=64+32\sqrt{13}+52=116+32\sqrt{13},$

so that is not correct. Try instead

$(a+b\sqrt{13})^2=a^2+13b^2+2ab\sqrt{13}.$

Matching $2ab=32$ and $a^2+13b^2=135$ gives $a=7$, $b=\frac{16}{7}$, which is not neat. So the radical is left as is unless the original problem intended a different root labeling.

Using the direct numerical forms of the roots gives

$P=\frac{\sqrt{135+32\sqrt{13}}+\frac{5-\sqrt{13}}{2}}{13}.$

If the roots are swapped, the same method applies, but the formula changes because $x_1$ appears inside the radical. The problem as stated depends on the labeling of the roots, so the exact value must respect the given order.

Pitfall alert

The biggest trap is assuming the expression is symmetric in $x_1$ and $x_2$. It is not, because $x_1$ appears inside the square root and $x_2$ appears added outside it. Another mistake is using Vieta's formulas to compute only the sum and product, then stopping too early when the expression still depends on one specific root. For such problems, you must respect the ordering of the roots or verify which root is which before simplifying.

Try different conditions

If the expression were changed to

$P'=\frac{\sqrt{64x_2-25}+x_1}{(x_1-x_2)^2},$

the same quadratic data would apply, but the value could be different because the root inside the square root has switched. A reliable strategy is to first determine which root is larger: for $x^2-5x+3=0$, the larger root is $\frac{5+\sqrt{13}}{2}$. Once the ordering is known, you can evaluate either variant consistently without re-solving the polynomial from scratch.

Further reading

Vieta's formulas, quadratic roots, discriminant