Question

1. $2x^2 + 7x = -3$

+3 +3
$2x^2 + 7x + 3 = 0$
$(x + 6)(x + 1) = 0$ p: 6, 6,1
s:7
$x = -6, -1$

1. $-2x = x^2 - 48$
2. $2x^2 - 11x + 5 = 0$
3. $6x^2 - 13x - 5 = 0$
4. $2x^2 + 5 = 11x$
5. $2x^2 - 32 = 0$

$x^2 - 16$
$(x + 4)(x - 4) = 0$
$x = -4, 4$

1. $x^2 - 3x + 2 = 0$
2. $x^2 - 5z + 4 = 0$
3. $x^2 - 8x + 16 = 0$

Explanation:

To solve these quadratic equations, we'll use methods like factoring, completing the square, or the quadratic formula. Let's solve a few of them:

Problem 43: \( 2x^2 + 7x = -3 \)

Step 1: Rewrite in standard form

Add 3 to both sides:
\( 2x^2 + 7x + 3 = 0 \)

Step 2: Factor the quadratic

We need two numbers that multiply to \( 2 \times 3 = 6 \) and add to 7. The numbers are 6 and 1.
Split the middle term:
\( 2x^2 + 6x + x + 3 = 0 \)
Factor by grouping:
\( 2x(x + 3) + 1(x + 3) = 0 \)
\( (2x + 1)(x + 3) = 0 \) (Note: The original factoring in the image was incorrect. Correct factoring is shown here.)

Step 3: Solve for \( x \)

Set each factor to zero:
\( 2x + 1 = 0 \implies x = -\frac{1}{2} \)
\( x + 3 = 0 \implies x = -3 \)

Problem 48: \( 2x^2 - 32 = 0 \)

Step 1: Factor out the GCF

Factor out 2:
\( 2(x^2 - 16) = 0 \)

Step 2: Factor the difference of squares

\( x^2 - 16 = (x + 4)(x - 4) \), so:
\( 2(x + 4)(x - 4) = 0 \)

Step 3: Solve for \( x \)

Set each factor to zero:
\( x + 4 = 0 \implies x = -4 \)
\( x - 4 = 0 \implies x = 4 \)

Problem 49: \( x^2 - 3x + 2 = 0 \)

Step 1: Factor the quadratic

We need two numbers that multiply to 2 and add to -3. The numbers are -1 and -2.
\( (x - 1)(x - 2) = 0 \)

Step 2: Solve for \( x \)

Set each factor to zero:
\( x - 1 = 0 \implies x = 1 \)
\( x - 2 = 0 \implies x = 2 \)

Problem 51: \( x^2 - 8x + 16 = 0 \)

Step 1: Factor the perfect square trinomial

\( x^2 - 8x + 16 = (x - 4)^2 \)

Step 2: Solve for \( x \)

\( (x - 4)^2 = 0 \implies x - 4 = 0 \implies x = 4 \) (double root)

Final Answers (for selected problems):

• 43: \( \boldsymbol{x = -\frac{1}{2}, -3} \)
• 48: \( \boldsymbol{x = -4, 4} \)
• 49: \( \boldsymbol{x = 1, 2} \)
• 51: \( \boldsymbol{x = 4} \)

Answer:

To solve these quadratic equations, we'll use methods like factoring, completing the square, or the quadratic formula. Let's solve a few of them:

Problem 43: \( 2x^2 + 7x = -3 \)

Step 1: Rewrite in standard form

Add 3 to both sides:
\( 2x^2 + 7x + 3 = 0 \)

Step 2: Factor the quadratic

We need two numbers that multiply to \( 2 \times 3 = 6 \) and add to 7. The numbers are 6 and 1.
Split the middle term:
\( 2x^2 + 6x + x + 3 = 0 \)
Factor by grouping:
\( 2x(x + 3) + 1(x + 3) = 0 \)
\( (2x + 1)(x + 3) = 0 \) (Note: The original factoring in the image was incorrect. Correct factoring is shown here.)

Step 3: Solve for \( x \)

Set each factor to zero:
\( 2x + 1 = 0 \implies x = -\frac{1}{2} \)
\( x + 3 = 0 \implies x = -3 \)

Problem 48: \( 2x^2 - 32 = 0 \)

Step 1: Factor out the GCF

Factor out 2:
\( 2(x^2 - 16) = 0 \)

Step 2: Factor the difference of squares

\( x^2 - 16 = (x + 4)(x - 4) \), so:
\( 2(x + 4)(x - 4) = 0 \)

Step 3: Solve for \( x \)

Set each factor to zero:
\( x + 4 = 0 \implies x = -4 \)
\( x - 4 = 0 \implies x = 4 \)

Problem 49: \( x^2 - 3x + 2 = 0 \)

Step 1: Factor the quadratic

We need two numbers that multiply to 2 and add to -3. The numbers are -1 and -2.
\( (x - 1)(x - 2) = 0 \)

Step 2: Solve for \( x \)

Set each factor to zero:
\( x - 1 = 0 \implies x = 1 \)
\( x - 2 = 0 \implies x = 2 \)

Problem 51: \( x^2 - 8x + 16 = 0 \)

Step 1: Factor the perfect square trinomial

\( x^2 - 8x + 16 = (x - 4)^2 \)

Step 2: Solve for \( x \)

\( (x - 4)^2 = 0 \implies x - 4 = 0 \implies x = 4 \) (double root)

Final Answers (for selected problems):

• 43: \( \boldsymbol{x = -\frac{1}{2}, -3} \)
• 48: \( \boldsymbol{x = -4, 4} \)
• 49: \( \boldsymbol{x = 1, 2} \)
• 51: \( \boldsymbol{x = 4} \)