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Quadratic Formula

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The Quadratic Formula

Solve the following quadratic equation algebraically:


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James Sousa: Using the Quadratic Formula

James Sousa: Proof of the Quadratic Formula


You can use the method of completing the square to solve the general quadratic equation ax^2+bx+c=0 . The result will be a formula that you can use to solve any quadratic equation given the values for a, b, and c . The following is a derivation of the quadratic formula:

Step 1: Divide the general equation by a . Then, move the third term on the left side to the right side of the equation.

& ax^2+bx+c=0\\& x^2+\frac{b}{{\color{red}a}}x+\frac{c}{{\color{red}a}}=0\\& x^2+\frac{b}{a}x=-\frac{c}{a}

Step 2: Complete the square. Note that your " b " value in this case is actually \frac{b}{a} .

& x^2+\frac{b}{a}x{\color{red}+\frac{b^2}{4a^2}}={\color{red}\frac{b^2}{4a^2}}-\frac{c}{a}

Step 3: Simplify.

& x^2+\frac{b}{a}x{\color{red}+\frac{b^2}{4a^2}}={\color{red}\frac{b^2}{4a^2}}-\frac{c}{a}\left(\frac{4a}{4a}\right)\\& x^2+\frac{b}{a}x{\color{red}+\frac{b^2}{4a^2}}={\color{red}\frac{b^2}{4a^2}}-\frac{4ac}{4a^2}\\& x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^2-4ac}{4a^2}

Step 4: Rewrite the left side of the equation as a binominal squared. Then, take the square root of both sides and solve for x .

&\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}\\&\sqrt{\left(x+\frac{b}{2a}\right)^2}=\sqrt{\frac{b^2-4ac}{4a^2}}\\& x+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}\\& x=-\frac{b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a} \\& x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

This is known as the quadratic formula. You can use the quadratic formula to solve ANY quadratic equation. All you need to know are the values of a, b, and c . Keep in mind that while the factoring method for solving a quadratic equation will only sometimes work, the quadratic formula will ALWAYS work. You should memorize the quadratic formula because you will use it in algebra and future math courses.

QUADRATIC FORMULA: x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Example A

Find the exact solutions of the following quadratic equation using the quadratic formula:


Solution: For this quadratic equation, a=5,b=2,c=-2 . Substitute these values into the quadratic formula and simplify.

& x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\& x=\frac{-({\color{red}2}) \pm \sqrt{({\color{red}2})^2-4({\color{red}5})({\color{red}-2})}}{2({\color{red}5})} \\& x=\frac{-2 \pm \sqrt{4+40}}{10}\\& x=\frac{-2 \pm \sqrt{44}}{10} \\& x =\frac{-2 \pm 2 \sqrt{11}}{10}\\& x=\frac{-1 \pm \sqrt{11}}{5}

The exact solutions to the quadratic equation are \frac{-1 + \sqrt{11}}{5} \ or \ \frac{-1-\sqrt{11}}{5} .

Example B

Use the quadratic formula to determine the approximate solutions of the equation:


Solution: Start by rewriting the equation in standard form so that it is set equal to zero. 2x^2-3x=3 becomes 2x^2-3x-3=0 . For this quadratic equation, a=2, b=-3, c=-3 . Substitute these values into the quadratic formula and simplify.

& m=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\& m=\frac{-({\color{red}-3}) \pm \sqrt{({\color{red}-3})^2-4({\color{red}2})({\color{red}-3})}}{2({\color{red}2})}\\& x=\frac{3 \pm \sqrt{9+24}}{4}\\& x=\frac{3 \pm \sqrt{33}}{4} \quad \sqrt{33}=5.74\\& x=\frac{3 \pm 5.74}{4} \\& x=\frac{3+5.74}{4} \ or \ x=\frac{3-5.74}{4}\\& x=\frac{8.74}{4} \ or \ x=\frac{-2.74}{4}\\& x=2.2 \ or \ x=-0.7

The approximate solutions to the quadratic equation to the nearest tenth are x=2.2 or x=-0.7 .

Example C

Solve the following equation using the quadratic formula:


Solution: While this does not look like a quadratic equation (it is actually a rational equation because it contains rational expressions), you can rewrite it as a quadratic equation by multiplying by (y)(y+1) to get rid of the fractions. Note that y and y+1 are the denominators you want to eliminate. This is why you want to multiply by (y)(y+1) . After multiplying, simplify and put the equation in standard quadratic form set equal to 0.

&\frac{2}{y}-\frac{3}{y+1}=1\\&\frac{2}{y}(y)(y+1)-\frac{3}{y+1}(y)(y+1)=1(y)(y+1)\\&\frac{2}{\cancel{y}}(\cancel{y})(y+1)-\frac{3}{\cancel{y+1}}(y)(\cancel{y+1})=1(y)(y+1)\\&2(y+1)-3(y)=1(y^2+y) \\&2y+2-3y=y^2+y\\&2-y=y^2+y\\& y^2+2y-2=0

For this quadratic equation, a=1,b=2,c=-2 . Substitute these values into the quadratic formula and simplify.

& y=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\& y=\frac{-({\color{red}2}) \pm \sqrt{({\color{red}2})^2-4({\color{red}1})({\color{red}-2})}}{2({\color{red}1})} \\& y=\frac{-2 \pm \sqrt{4+8}}{2}\\& y=\frac{-2 \pm \sqrt{12}}{2} \\& y=\frac{-2 \pm 2 \sqrt{3}}{2} \\& y=-1 \pm \sqrt{3}

The exact solutions to the equation are -1+\sqrt{3} or -1-\sqrt{3} . Note that neither of these solutions will cause the original equation to have a zero in the denominator, so they both work.

Concept Problem Revisited

To solve the equation 3x^2-5x+1=0 algebraically, you can use the quadratic formula. For this quadratic equation, a=3, b=-5, c=1 . Substitute these values into the quadratic formula and simplify.

& x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\& x=\frac{-({\color{red}-5}) \pm \sqrt{({\color{red}-5})^2-4({\color{red}3})({\color{red}1})}}{2({\color{red}3})} \\& x=\frac{5 \pm \sqrt{25-12}}{6}\\& x=\frac{5 \pm \sqrt{13}}{6}

The solutions are x=\frac{5 \pm \sqrt{13}}{6} .


Quadratic Formula
The quadratic formula is the formula \frac{-b \pm \sqrt{b^2-4ac}}{2a} used to determine the solutions for a quadratic equation.

Guided Practice

1. For the following equation, rewrite as a quadratic equation and state the values for a, b and c :


2. Solve the following quadratic equation using the quadratic formula:


3. Find the approximate solutions to the following equation:



1. Multiply by  (x-1)(x+2) to clear the fractions.

&\frac{2}{x-1}+\frac{3}{x+2}=1 \\&\frac{2}{x-1}(x-1)(x+2)+\frac{3}{x+2}(x-1)(x+2)=1(x-1)(x+2)\\&\frac{2}{\cancel{x-1}}(\cancel{x-1})(x+2)+\frac{3}{\cancel{x+2}}(x-1)({\cancel{x+2}})=1(x^2+2x-1x-2)\\&2(x+2)+3(x-1)=1(x^2+x-2) \\&2x+4+3x-3=x^2+x-2\\& x^2+x-2=5x+1 \\& x^2-4x-3=0

For this equation, a=1, b=-4, c=-3 .

2. This equation does not have a ‘ c ’ term. The value of ‘ c ’ is 0. For this equation, a=6, b=-8, c=0 .

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

x=\frac{-({\color{red}-8}) \pm \sqrt{({\color{red}-8})^2-4({\color{red}6})({\color{red}0})}}{2({\color{red}6})}

& x=\frac{8 \pm \sqrt{64-0}}{12}\\& x=\frac{8 \pm \sqrt{64}}{12} \\& x=\frac{-8 \pm 8}{12}\\& x=\frac{8+8}{12} \ or \ x=\frac{8-8}{12} \\&\boxed{x=\frac{4}{3} \ or \ x=0}

3. Multiply by (2x-1)(x+5) to clear the fractions.



For this equation, a=-3,b=4,c=18 .

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

x=\frac{-({\color{red}4}) \pm \sqrt{({\color{red}4})^2-4({\color{red}-3})({\color{red}18})}}{2({\color{red}-3})}

& x=\frac{-4 \pm \sqrt{16+216}}{-6}\\& x=\frac{-4 \pm \sqrt{232}}{-6} \\& x=3.21 \ and \ x=-1.87

The solutions to the quadratic equation to the nearest tenth are x=3.2 or x=-1.9 .


State the value of a, b and c for each of the following quadratic equations.

  1. 2x^2+7x-1=0
  2. 3x^2+2x=7
  3. 9x^2-7=4x
  4. 2x^2-7=0
  5. 4-2x^2=11x

Determine the exact roots of the following quadratic equations using the quadratic formula.

  1. 2x^2=8x-7
  2. 6y=2-y^2
  3. 1=8x+3x^2
  4. 2(n-2)(n+1)-(n+3)=0
  5. \frac{2e}{e+1}-\frac{3}{e-1}=\frac{4}{e^2-1}
  6. x^2-2x-5=0
  7. \frac{m}{4}-\frac{m^2}{2}=-1
  8. \frac{3}{y}-\frac{4}{y+2}=2
  9. \frac{1}{2}x^2-\frac{x}{4}-1=0
  10. 3x^2+8x=1




A binomial is an expression with two terms. The prefix 'bi' means 'two'.
Completing the Square

Completing the Square

Completing the square is a common method for rewriting quadratics. It refers to making a perfect square trinomial by adding the square of 1/2 of the coefficient of the x term.
Quadratic Formula

Quadratic Formula

The quadratic formula states that for any quadratic equation in the form ax^2+bx+c=0, x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}.


The roots of a function are the values of x that make y equal to zero.
Square Root

Square Root

The square root of a term is a value that must be multiplied by itself to equal the specified term. The square root of 9 is 3, since 3 * 3 = 9.


The vertex of a parabola is the highest or lowest point on the graph of a parabola. The vertex is the maximum point of a parabola that opens downward and the minimum point of a parabola that opens upward.

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