Question

①$7v^{2}-21=-8v$ ②$11n^{2}-7=8n$ ③$6p^{2}=-11p + 40$ ④$7x^{2}-5x + 21$

Explanation:

To solve these quadratic equations, we'll use the standard form \( ax^2 + bx + c = 0 \) and then apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) or factor if possible. Let's solve each equation one by one:

Equation ①: \( 7v^2 - 21 = -8v \)

Step 1: Rewrite in standard form

Add \( 8v \) to both sides:
\( 7v^2 + 8v - 21 = 0 \)
Here, \( a = 7 \), \( b = 8 \), \( c = -21 \).

Step 2: Apply the quadratic formula

\( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Substitute \( a = 7 \), \( b = 8 \), \( c = -21 \):
\( v = \frac{-8 \pm \sqrt{8^2 - 4(7)(-21)}}{2(7)} \)
Simplify the discriminant:
\( 8^2 - 4(7)(-21) = 64 + 588 = 652 \)
\( \sqrt{652} = \sqrt{4 \times 163} = 2\sqrt{163} \)
Thus:
\( v = \frac{-8 \pm 2\sqrt{163}}{14} = \frac{-4 \pm \sqrt{163}}{7} \)

Equation ②: \( 11n^2 - 7 = 8n \)

Step 1: Rewrite in standard form

Subtract \( 8n \) from both sides:
\( 11n^2 - 8n - 7 = 0 \)
Here, \( a = 11 \), \( b = -8 \), \( c = -7 \).

Step 2: Apply the quadratic formula

\( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Substitute \( a = 11 \), \( b = -8 \), \( c = -7 \):
\( n = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(11)(-7)}}{2(11)} \)
Simplify the discriminant:
\( (-8)^2 - 4(11)(-7) = 64 + 308 = 372 \)
\( \sqrt{372} = \sqrt{4 \times 93} = 2\sqrt{93} \)
Thus:
\( n = \frac{8 \pm 2\sqrt{93}}{22} = \frac{4 \pm \sqrt{93}}{11} \)

Equation ③: \( 6p^2 = -11p + 40 \)

Step 1: Rewrite in standard form

Add \( 11p - 40 \) to both sides:
\( 6p^2 + 11p - 40 = 0 \)
Here, \( a = 6 \), \( b = 11 \), \( c = -40 \).

Step 2: Factor (or use quadratic formula)

Let’s try factoring. We need two numbers that multiply to \( 6 \times (-40) = -240 \) and add to \( 11 \).
The numbers are \( 24 \) and \( -10 \) (since \( 24 \times (-10) = -240 \) and \( 24 + (-10) = 14 \)? Wait, no—wait, \( 24 \) and \( -10 \) sum to \( 14 \), not \( 11 \). Let’s use the quadratic formula instead.

\( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Substitute \( a = 6 \), \( b = 11 \), \( c = -40 \):
\( p = \frac{-11 \pm \sqrt{11^2 - 4(6)(-40)}}{2(6)} \)
Simplify the discriminant:
\( 11^2 - 4(6)(-40) = 121 + 960 = 1081 \)
\( \sqrt{1081} \) is irrational, so:
\( p = \frac{-11 \pm \sqrt{1081}}{12} \)

Equation ④: \( 7x^2 - 5x + 21 \)

This is not an equation (no “=”), so it’s an expression. If we assume it’s set to zero (\( 7x^2 - 5x + 21 = 0 \)):
\( a = 7 \), \( b = -5 \), \( c = 21 \)
Discriminant: \( (-5)^2 - 4(7)(21) = 25 - 588 = -563 \)
Since the discriminant is negative, there are no real solutions (only complex: \( x = \frac{5 \pm i\sqrt{563}}{14} \)).

Final Answers

• ① \( \boldsymbol{v = \frac{-4 \pm \sqrt{163}}{7}} \)
• ② \( \boldsymbol{n = \frac{4 \pm \sqrt{93}}{11}} \)
• ③ \( \boldsymbol{p = \frac{-11 \pm \sqrt{1081}}{12}} \)
• ④ (if \( 7x^2 - 5x + 21 = 0 \)): No real solutions (complex: \( \boldsymbol{x = \frac{5 \pm i\sqrt{563}}{14}} \))

Answer:

To solve these quadratic equations, we'll use the standard form \( ax^2 + bx + c = 0 \) and then apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) or factor if possible. Let's solve each equation one by one:

Equation ①: \( 7v^2 - 21 = -8v \)

Step 1: Rewrite in standard form

Add \( 8v \) to both sides:
\( 7v^2 + 8v - 21 = 0 \)
Here, \( a = 7 \), \( b = 8 \), \( c = -21 \).

Step 2: Apply the quadratic formula

\( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Substitute \( a = 7 \), \( b = 8 \), \( c = -21 \):
\( v = \frac{-8 \pm \sqrt{8^2 - 4(7)(-21)}}{2(7)} \)
Simplify the discriminant:
\( 8^2 - 4(7)(-21) = 64 + 588 = 652 \)
\( \sqrt{652} = \sqrt{4 \times 163} = 2\sqrt{163} \)
Thus:
\( v = \frac{-8 \pm 2\sqrt{163}}{14} = \frac{-4 \pm \sqrt{163}}{7} \)

Equation ②: \( 11n^2 - 7 = 8n \)

Step 1: Rewrite in standard form

Subtract \( 8n \) from both sides:
\( 11n^2 - 8n - 7 = 0 \)
Here, \( a = 11 \), \( b = -8 \), \( c = -7 \).

Step 2: Apply the quadratic formula

\( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Substitute \( a = 11 \), \( b = -8 \), \( c = -7 \):
\( n = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(11)(-7)}}{2(11)} \)
Simplify the discriminant:
\( (-8)^2 - 4(11)(-7) = 64 + 308 = 372 \)
\( \sqrt{372} = \sqrt{4 \times 93} = 2\sqrt{93} \)
Thus:
\( n = \frac{8 \pm 2\sqrt{93}}{22} = \frac{4 \pm \sqrt{93}}{11} \)

Equation ③: \( 6p^2 = -11p + 40 \)

Step 1: Rewrite in standard form

Add \( 11p - 40 \) to both sides:
\( 6p^2 + 11p - 40 = 0 \)
Here, \( a = 6 \), \( b = 11 \), \( c = -40 \).

Step 2: Factor (or use quadratic formula)

Let’s try factoring. We need two numbers that multiply to \( 6 \times (-40) = -240 \) and add to \( 11 \).
The numbers are \( 24 \) and \( -10 \) (since \( 24 \times (-10) = -240 \) and \( 24 + (-10) = 14 \)? Wait, no—wait, \( 24 \) and \( -10 \) sum to \( 14 \), not \( 11 \). Let’s use the quadratic formula instead.

\( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Substitute \( a = 6 \), \( b = 11 \), \( c = -40 \):
\( p = \frac{-11 \pm \sqrt{11^2 - 4(6)(-40)}}{2(6)} \)
Simplify the discriminant:
\( 11^2 - 4(6)(-40) = 121 + 960 = 1081 \)
\( \sqrt{1081} \) is irrational, so:
\( p = \frac{-11 \pm \sqrt{1081}}{12} \)

Equation ④: \( 7x^2 - 5x + 21 \)

This is not an equation (no “=”), so it’s an expression. If we assume it’s set to zero (\( 7x^2 - 5x + 21 = 0 \)):
\( a = 7 \), \( b = -5 \), \( c = 21 \)
Discriminant: \( (-5)^2 - 4(7)(21) = 25 - 588 = -563 \)
Since the discriminant is negative, there are no real solutions (only complex: \( x = \frac{5 \pm i\sqrt{563}}{14} \)).

Final Answers

• ① \( \boldsymbol{v = \frac{-4 \pm \sqrt{163}}{7}} \)
• ② \( \boldsymbol{n = \frac{4 \pm \sqrt{93}}{11}} \)
• ③ \( \boldsymbol{p = \frac{-11 \pm \sqrt{1081}}{12}} \)
• ④ (if \( 7x^2 - 5x + 21 = 0 \)): No real solutions (complex: \( \boldsymbol{x = \frac{5 \pm i\sqrt{563}}{14}} \))