Mejor AI Math Solver (2026): Herramientas probadas para obtener resultados precisos

### Steps

Given
\[
x+y=11,\qquad xy=24.
\]

Use
\[
x^2+y^2=(x+y)^2-2xy.
\]

Thus
\[
x^2+y^2=11^2-2(24)=121-48=73.
\]

To identify \(x\) and \(y\), they must be the roots of
\[
t^2-(x+y)t+xy=0,
\]
so
\[
t^2-11t+24=0.
\]

Factoring:
\[
(t-3)(t-8)=0.
\]

Hence the two values are \(3\) and \(8\).

### Final answer

\[
\boxed{x^2+y^2=73},\qquad \boxed{\{x,y\}=\{3,8\}}.
\]

### Verification

\[
3+8=11,\qquad 3\cdot 8=24,
\]
and
\[
3^2+8^2=9+64=73.
\]

Steps
- (x+y)² = x² + 2xy + y², so x² + y² = 11² �?? 2(24) = 121 �?? 48 = 73.
- x and y are roots of t² �?? 11t + 24 = 0 = (t �?? 3)(t �?? 8).

Final answer
x² + y² = 73; {x, y} = {3, 8}.

Verification
3 + 8 = 11 �??, 3·8 = 24 �??, 9 + 64 = 73 �??. Discriminant 121 �?? 96 = 25 > 0, so both roots are real.

Steps:
1. Use the algebraic identity $x^2 + y^2 = (x + y)^2 - 2xy$.
2. Substitute the given values $x + y = 11$ and $xy = 24$:
   $$x^2 + y^2 = 11^2 - 2(24) = 121 - 48 = 73$$
3. To find $x$ and $y$, construct the quadratic equation whose roots are $x$ and $y$:
   $$t^2 - (x+y)t + xy = 0 \implies t^2 - 11t + 24 = 0$$
4. Factor the quadratic equation:
   $$(t - 3)(t - 8) = 0 \implies t = 3 \text{ or } t = 8$$

Final answer:
$x^2 + y^2 = 73$; Unordered pair $\{x, y\} = \{3, 8\}$.

Verification:
$3 + 8 = 11$, $3 \times 8 = 24$, and $3^2 + 8^2 = 9 + 64 = 73$.

### Steps

Let the positive side lengths be \(a\) and \(b\).

The perimeter is \(34\) cm:
\[
2(a+b)=34,
\]
so
\[
a+b=17.
\]

The diagonal is \(13\) cm, so by the Pythagorean theorem,
\[
a^2+b^2=13^2=169.
\]

Using
\[
(a+b)^2=a^2+b^2+2ab,
\]
we obtain
\[
17^2=169+2ab.
\]

Therefore,
\[
289=169+2ab,
\]
\[
2ab=120,
\]
\[
ab=60.
\]

The side lengths are roots of
\[
t^2-17t+60=0.
\]

Factoring:
\[
(t-5)(t-12)=0.
\]

Thus the side lengths are \(5\) cm and \(12\) cm. The area is
\[
A=ab=60\text{ cm}^2.
\]

The dimensions are determined because their sum and product are fixed. The quadratic has exactly the two positive roots \(5\) and \(12\), which give the same rectangle up to interchanging length and width.

### Final answer

\[
\boxed{\text{Side lengths: }5\text{ cm and }12\text{ cm}}
\]
\[
\boxed{\text{Area: }60\text{ cm}^2}
\]

### Verification

Perimeter:
\[
2(5+12)=2(17)=34\text{ cm}.
\]

Diagonal:
\[
\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13\text{ cm}.
\]

Area:
\[
5\cdot 12=60\text{ cm}^2.
\]

Steps
- Let sides be a, b > 0. Perimeter: 2(a+b) = 34 �?? a + b = 17.
- Diagonal (Pythagoras): a² + b² = 169.
- 2ab = (a+b)² �?? (a²+b²) = 289 �?? 169 = 120 �?? ab = 60.
- a, b are roots of t² �?? 17t + 60 = 0 = (t �?? 5)(t �?? 12).

Final answer
Sides 5 cm and 12 cm; area 60 cm².

Verification
2(5+12) = 34 �??, 5² + 12² = 25 + 144 = 169 = 13² �??, area = 60 �??.
Why determined: the two conditions fix the sum and product of the side lengths, and a monic quadratic is uniquely determined by its sum and product of roots. Discriminant 289 �?? 240 = 49 > 0 gives two positive roots, which are the same rectangle up to relabeling a and b �?? so the dimensions are unique as an unordered pair.

Steps:
1. Let the side lengths be $a > 0$ and $b > 0$.
2. The perimeter equation is $2(a + b) = 34 \implies a + b = 17$.
3. By the Pythagorean theorem, the diagonal gives $a^2 + b^2 = 13^2 = 169$.
4. Use $(a + b)^2 = a^2 + b^2 + 2ab$:
   $$17^2 = 169 + 2ab \implies 289 = 169 + 2ab \implies 2ab = 120 \implies ab = 60$$
5. The area is $A = ab = 60\text{ cm}^2$.
6. Side lengths $a$ and $b$ are the positive roots of $t^2 - 17t + 60 = 0$:
   $$(t - 5)(t - 12) = 0 \implies t = 5 \text{ or } t = 12$$
7. **Why dimensions are determined:** The perimeter and diagonal uniquely fix both the sum $(a+b = 17)$ and product $(ab = 60)$ of the positive side lengths. A system specifying the sum and product of two numbers uniquely defines the set of roots of a quadratic polynomial, giving a unique pair of side lengths up to order.

Final answer:
Side lengths: $5\text{ cm}$ and $12\text{ cm}$; Area: $60\text{ cm}^2$. The dimensions are determined because perimeter and diagonal uniquely fix the sum and product of the adjacent side lengths, leading to a unique positive quadratic solution set.

Verification:
Perimeter: $2(5 + 12) = 34\text{ cm}$. Diagonal: $\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ cm}$. Area: $5 \times 12 = 60\text{ cm}^2$.

### Steps

Let the original cost be \(C\).

A \(25\%\) markup multiplies the cost by
\[
1+0.25=1.25.
\]

The marked price is therefore
\[
1.25C.
\]

A \(20\%\) discount means the customer pays \(80\%\) of the marked price:
\[
0.80(1.25C)=120.
\]

Since
\[
0.80\cdot 1.25=1,
\]
the equation becomes
\[
C=120.
\]

The paid price equals the original cost, so the net percentage change is
\[
\frac{120-120}{120}\cdot 100\%=0\%.
\]

### Final answer

\[
\boxed{\text{Original cost}=\$120}
\]
\[
\boxed{\text{Net percentage change}=0\%}
\]

### Verification

A \(25\%\) markup on \(\$120\) gives
\[
120(1.25)=\$150.
\]

A \(20\%\) discount on \(\$150\) gives
\[
150(0.80)=\$120.
\]

Thus the paid price is exactly equal to the original cost.

Steps
- Let cost = C. Marked price = 1.25C.
- Sale price = 0.80 �? 1.25C = 1.00C.
- Paid price = 120 �?? C = 120.
- Net change = (120 �?? 120)/120 = 0.

Final answer
Original cost = $120; net change from cost to paid price = 0% (no change).

Verification
120 �?? marked up 25% �?? 150 �?? 20% off �?? 150 �?? 30 = 120 �??. Note the multipliers: 1.25 �? 0.80 = 1 exactly, so the markup and discount cancel regardless of cost. (Result is before tax, as stated.)

Steps:
1. Let $C > 0$ be the original cost in dollars.
2. Markup of 25%: Marked price $M = C \times (1 + 0.25) = 1.25C$.
3. Discount of 20% on marked price: Paid price $P = M \times (1 - 0.20) = 1.25C \times 0.80 = 1.00C$.
4. Set $P = \$120$:
   $$1.00C = 120 \implies C = \$120$$
5. Net percentage change from cost to paid price:
   $$\text{Net Change \%} = \frac{P - C}{C} \times 100\% = \frac{120 - 120}{120} \times 100\% = 0\%$$

Final answer:
Original cost: $\$120$; Net percentage change: $0\%$.

Verification:
Markup: $\$120 \times 1.25 = \$150$. Discount: $\$150 \times 0.80 = \$120$. Paid price matches $\$120$, and cost equals paid price ($0\%$ net change).

### Steps

Solve
\[
\sqrt{x+6}=x
\]
over the real numbers.

Because a square root is nonnegative, the right-hand side must satisfy
\[
x\ge 0.
\]
Also, the radicand requires \(x+6\ge 0\), which is automatically satisfied when \(x\ge 0\).

Square both sides:
\[
x+6=x^2.
\]

Rearrange:
\[
x^2-x-6=0.
\]

Factor:
\[
(x-3)(x+2)=0.
\]

The candidates are
\[
x=3\quad\text{or}\quad x=-2.
\]

Both candidates must be checked in the original equation because squaring can introduce extraneous solutions.

For \(x=3\):
\[
\sqrt{3+6}=\sqrt9=3,
\]
so \(x=3\) is valid.

For \(x=-2\):
\[
\sqrt{-2+6}=\sqrt4=2\ne -2,
\]
so \(x=-2\) is invalid. It fails because the principal square root is always nonnegative, while the proposed right-hand side is negative.

### Final answer

\[
\boxed{x=3}
\]

### Verification

Substitution into the original equation gives
\[
\sqrt{3+6}=3,
\]
\[
3=3.
\]

The rejected candidate \(x=-2\) gives \(2=-2\), which is false.

Steps
- Domain/conditions: need x + 6 �?� 0 (x �?� �??6) and, since the principal square root is nonnegative, x �?� 0.
- Square both sides: x + 6 = x² �?? x² �?? x �?? 6 = 0 �?? (x �?? 3)(x + 2) = 0 �?? x = 3 or x = �??2.
- Test in the original equation: x = 3 �?? �??9 = 3 �??. x = �??2 �?? �??4 = 2 �?� �??2 �??.

Final answer
x = 3.

Verification
�??(3 + 6) = �??9 = 3 = x �??.
Why �??2 fails: squaring is not reversible �?? it turns �??(x+6) = x into an equation also satisfied by �??(x+6) = �??x. x = �??2 satisfies that second branch, but it violates the condition x �?� 0 because the principal root cannot be negative. It is an extraneous root introduced by squaring.

Steps:
1. Equation: $\sqrt{x + 6} = x$.
2. Domain/Condition Restrictions:
   - Radicand non-negative: $x + 6 \ge 0 \implies x \ge -6$.
   - Principal square root non-negative: $\sqrt{x + 6} \ge 0 \implies x \ge 0$.
   - Combined domain restriction: $x \ge 0$.
3. Square both sides:
   $$x + 6 = x^2 \implies x^2 - x - 6 = 0$$
4. Factor:
   $$(x - 3)(x + 2) = 0 \implies \text{Candidates: } x = 3, x = -2$$
5. Check candidates in original equation $\sqrt{x + 6} = x$:
   - For $x = 3$: $\sqrt{3 + 6} = \sqrt{9} = 3$. Valid (LHS = RHS).
   - For $x = -2$: $\sqrt{-2 + 6} = \sqrt{4} = 2 \neq -2$. Rejected.
6. **Reason for rejection:** The principal real square root function $\sqrt{\cdot}$ always yields a non-negative real value ($\sqrt{4} = +2$). Since the right side is $-2$, the equation $2 = -2$ is false. $x = -2$ is an extraneous solution introduced by squaring.

Final answer:
$x = 3$.

Verification:
LHS at $x = 3$: $\sqrt{3 + 6} = \sqrt{9} = 3 = \text{RHS}$.
LHS at $x = -2$: $\sqrt{-2 + 6} = \sqrt{4} = 2 \neq -2 = \text{RHS}$.

### Steps

Evaluate
\[
\int 2x e^{x^2}\,dx.
\]

Use \(u\)-substitution:
\[
u=x^2.
\]

Then
\[
du=2x\,dx.
\]

Therefore,
\[
\int 2x e^{x^2}\,dx=\int e^u\,du.
\]

Integrating:
\[
\int e^u\,du=e^u+C.
\]

Substitute \(u=x^2\):
\[
e^{x^2}+C.
\]

This antiderivative is valid for all real \(x\).

### Final answer

\[
\boxed{\int 2x e^{x^2}\,dx=e^{x^2}+C}
\]

### Verification

Differentiate the result using the chain rule:
\[
\frac{d}{dx}\left(e^{x^2}+C\right)
=e^{x^2}\cdot 2x
=2x e^{x^2}.
\]

This matches the original integrand.

Steps
- Substitution: u = x², du = 2x dx.
- �?� 2x e^(x²) dx = �?� e^u du = e^u + C = e^(x²) + C.

Final answer
e^(x²) + C.

Verification
d/dx [e^(x²) + C] = e^(x²) · d/dx(x²) = 2x e^(x²) �?? (chain rule). Valid for all real x; the integrand is continuous on �?�.

Steps:
1. Evaluate $\int 2x e^{x^2} \, dx$ over domain $x \in (-\infty, \infty)$.
2. Name of substitution: $u$-substitution with $u = x^2$.
3. Compute differential: $du = 2x \, dx$.
4. Substitute into integral:
   $$\int 2x e^{x^2} \, dx = \int e^u \, du = e^u + C$$
5. Back-substitute $u = x^2$:
   $$e^{x^2} + C$$

Final answer:
$e^{x^2} + C$  (Substitution: $u = x^2$)

Verification:
Differentiate the result with respect to $x$ using the chain rule:
$$\frac{d}{dx}\left(e^{x^2} + C\right) = e^{x^2} \cdot \frac{d}{dx}(x^2) + 0 = 2x e^{x^2}$$
This recovers the original integrand.