There are a number of ways to solve quadratics. Our ultimate goal when we say 'solve' is referring to finding the $x$x-intercepts or roots of the function. That is, the solutions of a quadratic equation relate to the $x$x-intercepts of the graph.
First we check for easy options:
Then we employ one of our solving methods:
When solving practical problems:
Solve the following equation: $\frac{10}{x}-3x=-1$10x−3x=−1.
Solution:
We can remove the fraction by multiplying every term by $x$x.
$10-3x^2=-x$10−3x2=−x
Take all the terms to one side so that a $0$0 appears in the equation (we may be able to factor later).
$3x^2-x-10=0$3x2−x−10=0
Using the Factor By Grouping Method, we need two numbers that have a product of $-30$−30 and sum of $-1$−1. The numbers that do this are $-6$−6 and $+5$+5.
$3x^2-x-10=0$3x2−x−10=0
Use the two numbers found to split $-x$−x into two terms.
$3x^2-6x+5x-10=0$3x2−6x+5x−10=0
We can now factor in pairs: $3x^2$3x2 and $-6x$−6x have common factors. $5x$5x and $-10$−10 have common factors.
$3x\left(x-2\right)+5\left(x-2\right)=0$3x(x−2)+5(x−2)=0
A common factor of $x-2$x−2 can be taken out.
$\left(3x+5\right)\left(x-2\right)=0$(3x+5)(x−2)=0
From here we can see that $x=2$x=2, $x=\frac{-5}{3}$x=−53.
Solve the equation $\frac{3-2x}{x}=4x$3−2xx=4x.
$3-2x=4x^2$3−2x=4x2 (Multiply both sides by $x$x)
$4x^2+2x-3=0$4x2+2x−3=0 (Rearrange to get all terms to one side of the equation)
Now, using the quadratic formula:
$x$x | $=$= | $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$−b±√b2−4ac2a |
$x$x | $=$= | $\frac{-2\pm\sqrt{2^2-4\times4\times\left(-3\right)}}{2\times4}$−2±√22−4×4×(−3)2×4 |
$x$x | $=$= | $\frac{-2\pm\sqrt{4+48}}{8}$−2±√4+488 |
$x$x | $=$= | $\frac{-2\pm\sqrt{52}}{8}$−2±√528 |
$x$x | $=$= | $\frac{-2\pm2\sqrt{13}}{8}$−2±2√138 |
$x$x | $=$= | $\frac{-1\pm\sqrt{13}}{4}$−1±√134 |
$x$x | $=$= | $\frac{-1}{4}\pm\frac{\sqrt{13}}{4}$−14±√134 |
So we have two distinct, irrational solutions, $x=\frac{-1+\sqrt{13}}{4}$x=−1+√134 and $x=\frac{-1-\sqrt{13}}{4}$x=−1−√134.
Sometimes we may be asked to solve an equation using a specific method, in which case that's the one we should use. But when no method is specified, how do we know which to use when we want to solve an equation of the form $ax^2+bx+c=0$ax2+bx+c=0?
In some cases it is easy to see that the equation can be factored and we wouldn't even think of using the quadratic formula. In other cases, especially when $a$a is not equal to $1$1, the factoring is not as easy to determine (using the Factor By Grouping Method or otherwise).
If it seems as though too much time is being spent trying to factor the equation, we can always achieve our goals using the quadratic formula. But if the quadratic formula gives rational solutions, we should be able to factor it!
Solve for $p$p: $5\left(p^2-3\right)-705=0$5(p2−3)−705=0 Write all solutions on the same line, separated by commas.
Solve for the unknown: $-8x+x^2=-6-x-x^2$−8x+x2=−6−x−x2 Write all solutions on the same line, separated by commas.
Solve the following equation for $x$x by substituting in $m=4^x$m=4x. $4^{2x}-65\times4^x+64=0$42x−65×4x+64=0 Use a comma to separate multiple solutions.