1. Solve for $$ x $$ if: $$ \begin{array}{ll}1.1 .1 & (2 x-1)(x+5)=0 \ 1.1 .2 & x(3 x+13)=11 \ 1.1 .3 & 1+\sqrt{2 x+1}=x \ 1.1 .4 & 2^{x \sqrt{x}}=2^{27} \ 1.1 .5 & x^{2}

Question

1. Solve for $$ x $$ if: $$ \begin{array}{ll}1.1 .1 & (2 x-1)(x+5)=0 \ 1.1 .2 & x(3 x+13)=11 \ 1.1 .3 & 1+\sqrt{2 x+1}=x \ 1.1 .4 & 2^{x \sqrt{x}}=2^{27} \ 1.1 .5 & x^{2}<-2 x+15 \ 	ext { Solve for } x & 	ext { and } y 	ext { simultaneously if: } \ 3-y+2 x=0 	ext { and } 6 x+4 y^{2}=3+5 x y\end{array} $$

UpStudy AI · Accepted Answer

**1.1.1. Solve $$ (2x-1)(x+5)=0 $$**

Since the product is zero, set each factor equal to zero:
$$
2x-1 = 0 \quad \text{or} \quad x+5=0.
$$
Solve each:
- For $$2x-1=0$$: 
  $$
  2x=1 \quad \Rightarrow \quad x=\frac{1}{2}.
  $$
- For $$x+5=0$$:
  $$
  x=-5.
  $$
Thus, the solutions are $$ x=\frac{1}{2} $$ and $$ x=-5 $$.

---

**1.1.2. Solve $$ x(3x+13)=11 $$**

Expand the left‐hand side:
$$
3x^2 + 13x = 11.
$$
Bring all terms to one side:
$$
3x^2 + 13x - 11 = 0.
$$
Use the quadratic formula:
$$
x=\frac{-13 \pm \sqrt{13^2-4\cdot3\cdot(-11)}}{2\cdot 3}.
$$
Compute the discriminant:
$$
13^2 -4\cdot3\cdot(-11)=169+132=301.
$$
Thus,
$$
x=\frac{-13 \pm \sqrt{301}}{6}.
$$

---

**1.1.3. Solve $$ 1+\sqrt{2x+1}=x $$**

Isolate the square root:
$$
\sqrt{2x+1} = x-1.
$$
**Note:** We require $$ x-1\ge0 $$ (i.e., $$ x\ge1 $$) and also $$ 2x+1\ge0 $$ (which holds for $$ x\ge -\frac{1}{2} $$). Thus, the primary restriction is $$ x\ge1 $$.

Square both sides:
$$
2x+1 = (x-1)^2.
$$
Expand the right-hand side:
$$
2x+1 = x^2 -2x+1.
$$
Rearrange to form a quadratic:
$$
0 = x^2 -2x -2x +1 -1 = x^2 -4x.
$$
Factor:
$$
x(x-4)=0.
$$
Thus, $$ x=0 $$ or $$ x=4 $$. However, $$ x=0 $$ does not satisfy $$ x\ge1 $$. Therefore, the only valid solution is $$ x=4 $$.

---

**1.1.4. Solve $$ 2^{x\sqrt{x}}=2^{27} $$**

Since the bases are the same, we equate the exponents:
$$
x\sqrt{x}=27.
$$
Express $$ \sqrt{x} $$ as $$ x^{1/2} $$:
$$
x \cdot x^{1/2}= x^{3/2} = 27.
$$
Solve for $$ x $$ by raising both sides to the power $$ \frac{2}{3} $$:
$$
x=27^{\frac{2}{3}}.
$$
Recognize that $$ 27^{\frac{1}{3}}=3 $$, thus:
$$
x=3^2=9.
$$

---

**1.1.5. Solve the inequality $$ x^2 < -2x+15 $$**

First, bring all terms to one side:
$$
x^2 + 2x -15 < 0.
$$
Find the roots of the quadratic equation $$ x^2+2x-15=0 $$ using the quadratic formula:
$$
x=\frac{-2 \pm \sqrt{2^2-4\cdot1\cdot(-15)}}{2\cdot1}=\frac{-2 \pm \sqrt{4+60}}{2}=\frac{-2 \pm \sqrt{64}}{2}=\frac{-2 \pm 8}{2}.
$$
Thus, the roots are:
$$
x = \frac{-2+8}{2}=3 \quad \text{and} \quad x = \frac{-2-8}{2}=-5.
$$
Since the quadratic coefficient is positive, the inequality holds between the roots:
$$
-5 < x < 3.
$$

---

**Simultaneous Equations**

Solve for $$ x $$ and $$ y $$ given:
$$
\begin{cases}
3-y+2x=0,\[1mm]
6x+4y^2=3+5xy.
\end{cases}
$$

**Step 1: Express $$ y $$ in terms of $$ x $$ from the first equation.**

$$
3-y+2x=0 \quad \Rightarrow \quad y=3+2x.
$$

**Step 2: Substitute $$ y=3+2x $$ into the second equation.**

$$
6x + 4(3+2x)^2 = 3 + 5x(3+2x).
$$

Calculate $$ (3+2x)^2 $$:
$$
(3+2x)^2=9+12x+4x^2.
$$
Thus, the equation becomes:
$$
6x+4(9+12x+4x^2)= 3+15x+10x^2.
$$
Simplify the left-hand side:
$$
6x + 36 + 48x + 16x^2 = 16x^2 + 54x + 36.
$$
So,
$$
16x^2 + 54x + 36 = 10x^2 + 15x+3.
$$

**Step 3: Bring all terms to one side.**

$$
16x^2 - 10x^2 + 54x - 15x + 36 - 3 = 0,
$$
$$
6x^2 + 39x + 33 = 0.
$$
Divide the entire equation by 3:
$$
2x^2 + 13x + 11 = 0.
$$

**Step 4: Solve the quadratic equation using the quadratic formula.**

$$
x=\frac{-13 \pm \sqrt{13^2-4\cdot2\cdot11}}{2\cdot2}=\frac{-13 \pm \sqrt{169-88}}{4}=\frac{-13 \pm \sqrt{81}}{4}=\frac{-13 \pm 9}{4}.
$$

Thus, the solutions for $$ x $$ are:
- $$ x=\frac{-13+9}{4}=\frac{-4}{4}=-1 $$,
- $$ x=\frac{-13-9}{4}=\frac{-22}{4}=-\frac{11}{2} $$.

**Step 5: Find $$ y $$ using $$ y= 3+2x $$.**

For $$ x=-1 $$:
$$
y=3+2(-1)=3-2=1.
$$

For $$ x=-\frac{11}{2} $$:
$$
y=3+2\left(-\frac{11}{2}\right)=3-11=-8.
$$

Thus, the solutions are:
$$
(x,y)=(-1,\,1) \quad \text{and} \quad \left(-\frac{11}{2},\,-8\right).
$$
**Solutions:** 1. **1.1.1:** $$ x = \frac{1}{2} $$ or $$ x = -5 $$ 2. **1.1.2:** $$ x = \frac{-13 \pm \sqrt{301}}{6} $$ 3. **1.1.3:** $$ x = 4 $$ 4. **1.1.4:** $$ x = 9 $$ 5. **1.1.5:** $$ -5 < x < 3 $$ **Simultaneous Equations:** - $$ (x, y) = (-1, 1) $$ - $$ \left(-\frac{11}{2}, -8\right) $$

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