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In the previous lesson, you learned a formula for solving quadratic equations in vertex form. You also learned how to convert quadratic equations from standard form to vertex form, by completing the square. In this lesson, you will combine these two techniques to obtain a general formula for solving any quadratic equation in standard form.

First, remember that you can simplify square roots by factoring out a perfect square, and using the fact that $√{ab}=√a√b$ whenever $a ≥ 0$ and $b ≥ 0$. For example, $√12=√{4(3)}=√4√3=2√3$.

Simplify the following square roots, by factoring out the largest possible perfect square.

Similarly, you can simplify certain square roots of fractions by using the fact that $$√{a/b}={√a}/{√b}$$ whenever $a ≥ 0$ and $b > 0$.

Simplify the following square roots of fractions.

Solve the equation $3(x-2)^2-5=0$, by using the formula $$x=h ± √{-k/a}$$ for equations in vertex form ($a(x-h)^2+k=0$). |

How many solutions does the equation $3(x-2)^2-5=0$ have? |

Now, we will solve $2x^2-5x+3=0$ by rewriting it in vertex form.

Rewrite $2x^2-5x+3$ in vertex form, using the formulas $$h=-b/{2a}$$ and $$k=-ah^2+c$$. (Remember, the value of $a$ does not change between standard and vertex form.)

What is the vertex form of the equation $2x^2-5x+3=0$? |

What are the solutions to that equation?

Given any equation of the form $a(x-h)^2+k=0$ with $a≠0$, we found its solutions are $$x = h ± √{{-k}/a}$$. We want to find a similar formula to solve any equation of the form $ax^2+bx+c=0$ with $a≠0$.

Given such an equation, we start by completing the square. From the last lesson, we know that $ax^2+bx+c = a(x-h)^2+k$ if $$h=-b/{2a}$$ and $k=-ah^2+c$. We can rewrite $k$ as follows:

$$k=-ah^2+c=-a(-b/{2a})^2+c=-a({b^2}/{4a^2})+c=-{b^2}/{4a}+c$$We want to use this to rewrite $${-k}/a$$ (the expression inside the square root). We’ll start by rewriting $-k$, as follows:

$$-k = -(-{b^2}/{4a}+c) = {b^2}/{4a} - c$$Now, using a common denominator for this last expression gives

$$ -k = {b^2}/{4a} - {4ac}/{4a} = {b^2-4ac}/{4a} $$and therefore

$${-k}/a = {b^2-4ac}/{4a} ∕ a = ({b^2-4ac}/{4a})(1/a) = {b^2-4ac}/{4a^2}$$So the equation $ax^2+bx+c=0$ has the solutions

$$x = h ± √{{-k}/a} = -b/{2a} ± √{{b^2-4ac}/{4a^2}} = -b/{2a} ± {√{b^2-4ac}}/{√{4a^2}}$$which simplfies to

$$x = -b/{2a} ± {√{b^2-4ac}}/{2a} = {-b ± √{b^2-4ac}}/{2a}$$This is the formula we were looking for, called the quadratic formula.

If $a ≠ 0$, the equation $ax^2+bx+c=0$ has the solutions $$x = {-b ± √{b^2-4ac}} / {2a}$$ whenever $b^2-4ac ≥ 0$. The equation has two solutions when $b^2-4ac > 0$; it has one solution when $b^2-4ac=0$; and it has no real solutions when $b^2-4ac < 0$.

Use the quadratic formula to find the solutions to the equation $2x^2-5x+3=0$.

How many solutions does the equation $2x^2-5x+3=0$ have? |

Are these the same solutions you found in compSquareSolns? |

Use the quadratic formula to find the solutions to the equation $2x^2-5x+1=0$.

How many solutions does the equation $2x^2-5x+1=0$ have? |

Use the quadratic formula to solve the equation $x^2+4x+1=0$.

How many solutions does the equation $x^2+4x+1=0$ have? |

Use the fact that $√{ab}=√a√b$ to simplify the solutions you found, by factoring the largest possible square number out of the square root.

Now, we would like to check that the simplified solutions we found at the end of simpQuad — namely, $x=-2 ± √3$ — solve the original equation $x^2+4x+1=0$.

$√3$ is approximately $1.732$. Look at the graph to the left. Do the numbers $x = -2 ± 1.732$ appear to be approximate solutions to the equation $x^2+4x+1=0$? |

To check that $x=-2+√3$ is an exact solution, first we square it:

$$ \cl"tight"{\table , \colspan 7 (-2+√3)^2; =, \colspan 7 (-2+√3)(-2+√3); =, \colspan 3 (-2)(-2+√3), +, \colspan 3 √3(-2+√3); =, (-2)(-2), +, (-2)√3, +, √3(-2), +, √3√3; =, 4, -, 2√3, -, 2√3, +, 3; =, \colspan 7 7-4√3 } $$and then we plug both $(-2+√3)$ and its square $(7-4√3)$ back into the original equation:

$$ \cl"tight"{\table \colspan 3 x^2, +, \colspan 3 4x, +, 1, = , 0; \colspan 3 (7 - 4√3), +, \colspan 3 4(-2 + √3), +, 1, =, 0; 7, -, 4√3, -, 8, +, 4√3, +, 1, =, 0; \colspan 4 7-8+1, \colspan 5 -4√3+4√3, =, 0; \colspan 9 0, =,0 } $$Find the square of $(-2-√3)$. (You may want to compare your work to the example above where we found the square of $(-2+√3)$.)

Check that $x=-2-√3$ is a solution to the equation $x^2+4x+1=0$, by plugging $(-2-√3)$ and its square into that equation.

All of the quadratic equations we’ve looked at up until now have had two different solutions. This is because the quantity $b^2-4ac$ — which appears under the square root in the quadratic formula — has always been positive. In this section, we’ll look at what happens to the number of solutions when that quantity is no longer positive.

We’ll begin by looking at the equation $x^2+2x+2=0$. What is the quantity $b^2-4ac$ for this equation?

For this equation, does the quantity $b^2-4ac$ have a real square root? |

Does this equation have any real solutions? |

Now, we’ll look at the equation $9x^2+6x+1=0$. What is the quantity $b^2-4ac$ for this equation?

Use the quadratic formula to solve this equation. Remember that you’ve already computed the quantity $b^2-4ac$, which goes under the square root in the quadratic formula.

How many solutions does the equation $9x^2+6x+1=0$ have? |

The expression $b^2-4ac$ appears under a square root in the quadratic formula. This means that it tells you what the solutions to the quadratic equation $ax^2+bx+c=0$ will look like.

The expression $b^2-4ac$ is called the discriminant of the equation $ax^2+bx+c=0$. The equation has two solutions if its discriminant is positive, one solution if its discriminant is zero, and no real solutions if its discriminant is negative.

Compute the discriminant of each quadratic equation in the table below, and give the number of real solutions to that equation.

Equation | Discriminant | Number of solutions |
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Because you’re taking the square root of the discriminant in the quadratic formula, it also tells you whether the solutions to the equation $ax^2+bx+c=0$ will be rational numbers or not.

If $a$, $b$, and $c$ are rational numbers with $a ≠ 0$, then the solutions to the equation $ax^2+bx+c=0$ are rational numbers whenever its discriminant $b^2-4ac$ is a perfect square.

Compute the discriminant of each quadratic equation in the table below, and indicate whether its solutions will be rational numbers.

Equation | Discriminant | Are the solutions rational? |
---|

The quadratic formula allows you to solve quadratic equations of the form $ax^2+bx+c=0$. As you’ve previously learned, any equation where both sides are quadratic polynomials can be written in this form, by subtracting the right-hand side from both sides of the equation.

Using scratch paper, rewrite each equation in the form $ax^2+bx+c=0$. Then solve it by using the quadratic formula. (Remember to reduce any fractions to lowest terms.)

Similarly, you can solve any equation where both sides simplify into quadratic polynomials.

We would like to solve the equation $x(2x-7)=-3(x-1)^2+9$. First, use scratch paper to rewrite each side of this equation in the form $ax^2+bx+c$.

Now, rewrite the entire equation in the form $ax^2+bx+c=0$, as in generalQuadQn.

Finally, use the quadratic formula to solve this simplified equation.