Solve for $$ x $$ : $$ \begin{array}{ll} ext { (1) } 4 x^{2}=9 x & ext { (2) } x^{2}=25 \ ext { (3) } x^{2}-5 x=14 & ext { (4) }-x^{2}-4 x+12=0 \ ext { (5) } 2 x^{2}-5 x+3=0 & ext { (6) } 6 x^{2}+7 x=20 \ ext { (7) }(x-8)(x+2)=-24 & ext { (8) } \ ext { (2x-3) }(x+1)=-2\end{array} $$

Question

Solve for $$ x $$ : $$ \begin{array}{ll}	ext { (1) } 4 x^{2}=9 x & 	ext { (2) } x^{2}=25 \ 	ext { (3) } x^{2}-5 x=14 & 	ext { (4) }-x^{2}-4 x+12=0 \ 	ext { (5) } 2 x^{2}-5 x+3=0 & 	ext { (6) } 6 x^{2}+7 x=20 \ 	ext { (7) }(x-8)(x+2)=-24 & 	ext { (8) } \ 	ext { (2x-3) }(x+1)=-2\end{array} $$

UpStudy AI · Accepted Answer

**(1)** Solve $$ 4x^2 = 9x $$.  
Rearrange the equation:
$$
4x^2 - 9x = 0
$$
Factor out $$ x $$:
$$
x(4x - 9) = 0
$$
Set each factor equal to zero:
$$
x = 0 \quad \text{or} \quad 4x - 9 = 0
$$
Solving $$ 4x - 9 = 0 $$:
$$
4x = 9 \quad \Longrightarrow \quad x = \frac{9}{4}
$$
Thus, the solutions are:
$$
x = 0 \quad \text{or} \quad x = \frac{9}{4}
$$

---

**(2)** Solve $$ x^2 = 25 $$.  
Take the square root of both sides:
$$
x = \sqrt{25} \quad \text{or} \quad x = -\sqrt{25}
$$
Thus:
$$
x = 5 \quad \text{or} \quad x = -5
$$

---

**(3)** Solve $$ x^2 - 5x = 14 $$.  
Rearrange the equation:
$$
x^2 - 5x - 14 = 0
$$
Factor the quadratic. We look for two numbers that multiply to $$-14$$ and add to $$-5$$. These numbers are $$-7$$ and $$2$$:
$$
(x - 7)(x + 2) = 0
$$
Thus:
$$
x = 7 \quad \text{or} \quad x = -2
$$

---

**(4)** Solve $$ -x^2 - 4x + 12 = 0 $$.  
Multiply the entire equation by $$-1$$ to simplify:
$$
x^2 + 4x - 12 = 0
$$
Factor the quadratic. We need two numbers that multiply to $$-12$$ and add to $$4$$; these are $$6$$ and $$-2$$:
$$
(x + 6)(x - 2) = 0
$$
Thus:
$$
x = -6 \quad \text{or} \quad x = 2
$$

---

**(5)** Solve $$ 2x^2 - 5x + 3 = 0 $$.  
Use the quadratic formula:
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
Here, $$a = 2$$, $$b = -5$$, and $$c = 3$$. Compute the discriminant:
$$
b^2 - 4ac = (-5)^2 - 4(2)(3) = 25 - 24 = 1
$$
Thus:
$$
x = \frac{5 \pm \sqrt{1}}{4} = \frac{5 \pm 1}{4}
$$
So,
$$
x = \frac{6}{4} = \frac{3}{2} \quad \text{or} \quad x = \frac{4}{4} = 1
$$

---

**(6)** Solve $$ 6x^2 + 7x = 20 $$.  
Rearrange the equation:
$$
6x^2 + 7x - 20 = 0
$$
Use the quadratic formula with $$a = 6$$, $$b = 7$$, and $$c = -20$$:
$$
x = \frac{-7 \pm \sqrt{7^2 - 4(6)(-20)}}{12}
$$
Calculate the discriminant:
$$
7^2 - 4(6)(-20) = 49 + 480 = 529
$$
Since $$ \sqrt{529} = 23 $$, we have:
$$
x = \frac{-7 \pm 23}{12}
$$
Thus,
$$
x = \frac{-7 + 23}{12} = \frac{16}{12} = \frac{4}{3} \quad \text{or} \quad x = \frac{-7 - 23}{12} = \frac{-30}{12} = -\frac{5}{2}
$$

---

**(7)** Solve $$ (x-8)(x+2) = -24 $$.  
Expand the left-hand side:
$$
x^2 - 8x + 2x - 16 = x^2 - 6x - 16
$$
Thus, the equation is:
$$
x^2 - 6x - 16 = -24
$$
Add $$ 24 $$ to both sides:
$$
x^2 - 6x + 8 = 0
$$
Factor the quadratic:
$$
(x - 2)(x - 4) = 0
$$
Thus:
$$
x = 2 \quad \text{or} \quad x = 4
$$

---

**(8)** Solve $$ (2x-3)(x+1) = -2 $$.  
Expand the left-hand side:
$$
2x^2 + 2x - 3x - 3 = 2x^2 - x - 3
$$
Thus, the equation becomes:
$$
2x^2 - x - 3 = -2
$$
Add $$2$$ to both sides:
$$
2x^2 - x - 1 = 0
$$
Apply the quadratic formula with $$a = 2$$, $$b = -1$$, and $$c = -1$$:
$$
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm \sqrt{9}}{4}
$$
Since $$\sqrt{9} = 3$$, we have:
$$
x = \frac{1 \pm 3}{4}
$$
Thus,
$$
x = \frac{1+3}{4} = 1 \quad \text{or} \quad x = \frac{1-3}{4} = -\frac{1}{2}
$$
Here are the solutions for each equation: 1. $$ x = 0 $$ or $$ x = \frac{9}{4} $$ 2. $$ x = 5 $$ or $$ x = -5 $$ 3. $$ x = 7 $$ or $$ x = -2 $$ 4. $$ x = -6 $$ or $$ x = 2 $$ 5. $$ x = \frac{3}{2} $$ or $$ x = 1 $$ 6. $$ x = \frac{4}{3} $$ or $$ x = -\frac{5}{2} $$ 7. $$ x = 2 $$ or $$ x = 4 $$ 8. $$ x = 1 $$ or $$ x = -\frac{1}{2} $$

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