Which equation shows the quadratic formula used correctly to solve \( 7x^2 = 9 + x \) for \( x \)?

Which equation shows the quadratic formula used correctly to solve \( 7x^2 = 9 + x \) for \( x \)?

December 15, 2024

Question: Which equation shows the quadratic formula used correctly to solve 7x2=9+x7x^2 = 9 + x for xx?

Answer: . x=1±124(7)(9)2(7)x = \frac{-1 \pm \sqrt{1^2 - 4(7)(9)}}{2(7)}

Explanation:

Step 1: Rearranging the equation into standard quadratic form

The given equation is 7x2=9+x7x^2 = 9 + x. To use the quadratic formula, we first need to rewrite it in the form ax2+bx+c=0ax^2 + bx + c = 0.

7x2x9=07x^2 - x - 9 = 0

Step 2: Identifying the coefficients

Now the quadratic equation is in the form ax2+bx+c=0ax^2 + bx + c = 0, where:

  • a=7a = 7
  • b=1b = -1
  • c=9c = -9

Step 3: Applying the quadratic formula

The quadratic formula is given by:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting the values of aa, bb, and cc into the formula:

x=(1)±(1)24(7)(9)2(7)x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(7)(-9)}}{2(7)}

This simplifies to:

x=1±1+25214x = \frac{1 \pm \sqrt{1 + 252}}{14} x=1±25314x = \frac{1 \pm \sqrt{253}}{14}

Thus, the correct form of the equation is:

x=1±124(7)(9)2(7)x = \frac{-1 \pm \sqrt{1^2 - 4(7)(9)}}{2(7)}

Extended Knowledge:

Quadratic Formula

The quadratic formula is a method for solving any quadratic equation of the form ax2+bx+c=0ax^2 + bx + c = 0. It is derived from completing the square of the equation and provides the solutions for xx. The two possible solutions are given by the ±\pm symbol, indicating two possible values for xx.

Completing the Square

Completing the square is a method used to solve quadratic equations by rewriting the equation in a perfect square form. This method ultimately leads to the quadratic formula and can be an alternative way to find the roots of a quadratic equation.


Similar Questions

Question 1: Solve the quadratic equation 3x26x+2=03x^2 - 6x + 2 = 0 using the quadratic formula.

Answer:

x=(6)±(6)24(3)(2)2(3)x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(2)}}{2(3)}

Explanation:

Step 1: Identify the coefficients

For the equation 3x26x+2=03x^2 - 6x + 2 = 0, we have:

  • a=3a = 3
  • b=6b = -6
  • c=2c = 2

Step 2: Apply the quadratic formula

The quadratic formula is:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting the values of aa, bb, and cc into the formula:

x=(6)±(6)24(3)(2)2(3)=6±36246=6±126x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(2)}}{2(3)} = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm \sqrt{12}}{6}

Step 3: Simplify the result

x=6±236=1±33x = \frac{6 \pm 2\sqrt{3}}{6} = 1 \pm \frac{\sqrt{3}}{3}

Thus, the solutions for xx are:

x=1+33,x=133x = 1 + \frac{\sqrt{3}}{3}, \quad x = 1 - \frac{\sqrt{3}}{3}

Extended Knowledge:

Quadratic Formula

The quadratic formula x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} is used to solve any quadratic equation of the form ax2+bx+c=0ax^2 + bx + c = 0. It can provide both real and complex roots depending on the discriminant (b24acb^2 - 4ac).


Question 2: Solve the quadratic equation 4y2+8y5=04y^2 + 8y - 5 = 0 using the quadratic formula.

Answer:

y=8±824(4)(5)2(4)y = \frac{-8 \pm \sqrt{8^2 - 4(4)(-5)}}{2(4)}

Explanation:

Step 1: Identify the coefficients

For the equation 4y2+8y5=04y^2 + 8y - 5 = 0, we have:

  • a=4a = 4
  • b=8b = 8
  • c=5c = -5

Step 2: Apply the quadratic formula

y=8±824(4)(5)2(4)=8±64+808=8±1448y = \frac{-8 \pm \sqrt{8^2 - 4(4)(-5)}}{2(4)} = \frac{-8 \pm \sqrt{64 + 80}}{8} = \frac{-8 \pm \sqrt{144}}{8}

Step 3: Simplify the result

y=8±128y = \frac{-8 \pm 12}{8}

Thus, the solutions for yy are:

y=8+128=48=12,y=8128=208=52y = \frac{-8 + 12}{8} = \frac{4}{8} = \frac{1}{2}, \quad y = \frac{-8 - 12}{8} = \frac{-20}{8} = -\frac{5}{2}

Question 3: Solve the quadratic equation 2z23z+1=02z^2 - 3z + 1 = 0 using the quadratic formula.

Answer:

z=(3)±(3)24(2)(1)2(2)z = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(1)}}{2(2)}

Explanation:

Step 1: Identify the coefficients

For the equation 2z23z+1=02z^2 - 3z + 1 = 0, we have:

  • a=2a = 2
  • b=3b = -3
  • c=1c = 1

Step 2: Apply the quadratic formula

z=(3)±(3)24(2)(1)2(2)=3±984=3±14z = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(1)}}{2(2)} = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm \sqrt{1}}{4}

Step 3: Simplify the result

z=3±14z = \frac{3 \pm 1}{4}

Thus, the solutions for zz are:

z=3+14=44=1,z=314=24=12z = \frac{3 + 1}{4} = \frac{4}{4} = 1, \quad z = \frac{3 - 1}{4} = \frac{2}{4} = \frac{1}{2}

Question 4: Solve the quadratic equation 5t2+7t3=05t^2 + 7t - 3 = 0 using the quadratic formula.

Answer:

t=7±724(5)(3)2(5)t = \frac{-7 \pm \sqrt{7^2 - 4(5)(-3)}}{2(5)}

Explanation:

Step 1: Identify the coefficients

For the equation 5t2+7t3=05t^2 + 7t - 3 = 0, we have:

  • a=5a = 5
  • b=7b = 7
  • c=3c = -3

Step 2: Apply the quadratic formula

t=7±724(5)(3)2(5)=7±49+6010=7±10910t = \frac{-7 \pm \sqrt{7^2 - 4(5)(-3)}}{2(5)} = \frac{-7 \pm \sqrt{49 + 60}}{10} = \frac{-7 \pm \sqrt{109}}{10}

Thus, the solutions for tt are:

t=7+10910,t=710910t = \frac{-7 + \sqrt{109}}{10}, \quad t = \frac{-7 - \sqrt{109}}{10}