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Solving x² = c and using formulas
Solving equations of the form x² = c with two, one or no solutions, and substituting into formulas to find an unknown.
Equations with $x^2$
Both $5^2$ and $(-5)^2$ equal $25$, so $x^2 = 25$ has two solutions: $x = \pm 5$.
Move the line up and down. Each crossing point is a solution.
Solve $2x^2 - 8 = 42$.
Using formulas
Use $v = u + at$ to find $a$ when $v = 30$, $u = 10$ and $t = 4$.
Solve $2x^2 - 8 = 42$. Enter both solutions separated by a comma.
$x^2 = 25$.
$v = u + at$. Find $a$ when $v = 30$, $u = 10$, $t = 4$.
$4a = 20$.
How many solutions does $x^2 = -9$ have?
A square cannot be negative.
Solve $x^2 = 49$. Enter both solutions.
$\pm\sqrt{49}$.
Solve $x^2 + 9 = 25$.
$x^2 = 16$.
$A = lw$. Find $w$ when $A = 56$ and $l = 7$.
$56 \div 7$.
$P = 2l + 2w$. Find $l$ when $P = 40$ and $w = 8$.
$2l = 24$.
$A = \tfrac{1}{2}bh$. Find $h$ when $A = 24$ and $b = 8$.
$24 = 4h$.
New numbers every time. Keep going until you get five in a row.