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Solving simple quadratics review

Simple quadratic equations like x^2=4 can be solved by taking the square root. This article reviews several examples and gives you a chance to practice on your own.
In general, a quadratic equation can be written as:
a, x, squared, plus, b, x, plus, c, equals, 0
In this article, we review how to solve quadratics that are solvable by taking the square root—no fancy factoring or quadratic equations here; we'll get to that technique later.

Example 1

We're given 3, x, squared, minus, 7, equals, 5 and asked to solve for x.
We can show our work like this:
3x27=53x2=12x2=4x2=±4x=±2\begin{aligned} 3x^2-7&=5\\\\ 3x^2&=12\\\\ x^2&=4\\\\ \sqrt{x^2}&=\pm \sqrt{4}\\\\ x&=\pm 2 \end{aligned}
So our two solutions are:
  • x, equals, 2
  • x, equals, minus, 2
Notice the plus minus symbol we included when taking the square root of both sides. This symbol means "plus or minus," and it is important because it ensures we catch both solutions. Want a deeper explanation? Check out this video.
Let's check both solutions:
x, equals, 2x, equals, minus, 2
3x27=53(2)27=5347=5127=55=5\begin{aligned}3x^2-7&=5\\\\3(2)^2-7&=5\\\\3\cdot4-7&=5\\\\12-7&=5\\\\5&=5\end{aligned}3x27=53(2)27=5347=5127=55=5\begin{aligned}3x^2-7&=5\\\\3(-2)^2-7&=5\\\\3\cdot4-7&=5\\\\12-7&=5\\\\5&=5\end{aligned}
Yes! Both solutions check out.

Example 2

We're given left parenthesis, x, minus, 3, right parenthesis, squared, minus, 81, equals, 0 and asked to solve for x.
We can show our work like this:
(x3)281=0(x3)2=81(x3)2=±81x3=±9x=±9+3\begin{aligned} (x - 3)^2 - 81 &= 0\\\\ (x - 3)^2 &= 81\\\\ \sqrt{(x - 3)^2} &= \pm \sqrt{81}\\\\ x - 3 &= \pm 9\\\\ x &= \pm 9+3 \end{aligned}
So our two solutions are:
  • x, equals, plus, 9, plus, 3, equals, start color #11accd, 12, end color #11accd
  • x, equals, minus, 9, plus, 3, equals, start color #11accd, minus, 6, end color #11accd
Let's check both solutions:
x, equals, start color #11accd, 12, end color #11accdx, equals, start color #11accd, minus, 6, end color #11accd
(x3)281=0(123)281=09281=08181=00=0\begin{aligned}(x - 3)^2 - 81 &= 0\\\\(\blueD{12} - 3)^2 - 81 &= 0\\\\9^2 - 81 &= 0\\\\81 - 81 &= 0\\\\0 &= 0\end{aligned}(x3)281=0(63)281=0(9)281=08181=00=0\begin{aligned}(x - 3)^2 - 81 &= 0\\\\(\blueD{-6} - 3)^2 - 81 &= 0\\\\(-9)^2 - 81 &= 0\\\\81 - 81 &= 0\\\\0 &= 0\end{aligned}
Yep! Both check out.
Want to learn more about these types of problems? Check out this video.
Practice
Solve for x.
left parenthesis, x, plus, 1, right parenthesis, squared, minus, 36, equals, 0
Choose 1 answer:
Choose 1 answer:

Want more practice? Check out this exercise

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