$\sqrt{7-3 z}=3+c z$

View step-by-step help

Home

Math Problems

Grade 7

Powers with negative bases

Full solution

\sqrt{7-3 z}=3+c z

Isolate square root: First, we need to isolate the square root on one side of the equation. The equation is already in the form we need: $\sqrt{7-3z}$ on one side and $3+cz$ on the other.
Square both sides: Next, we square both sides of the equation to eliminate the square root. This gives us $(\sqrt{7-3z})^2 = (3+cz)^2$ .
Apply binomial theorem: Squaring the left side, we get $7-3z$ . Squaring the right side, we need to apply the binomial theorem $(a+b)^2 = a^2 + 2ab + b^2$ , which gives us $9 + 6cz + c^2z^2$ .
Set equation to zero: Now we have the equation $7-3z = 9 + 6cz + c^2z^2$ . To solve for $z$ , we need to set the equation to zero by moving all terms to one side: $0 = c^2z^2 + 6cz + (9 - 7)$ .
Simplify constant terms: Simplify the constant terms on the right side of the equation: $0 = c^2z^2 + 6cz + 2$ .
Quadratic equation in $z$ : We now have a quadratic equation in terms of $z$ : $c^2z^2 + 6cz + 2 = 0$ . To solve for $z$ , we can use the quadratic formula $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = c^2$ , $b = 6c$ , and $c = 2$ .
Substitute values: Substitute the values into the quadratic formula: $z = \frac{-(6c) \pm \sqrt{(6c)^2 - 4(c^2)(2)}}{2c^2}$ .
Simplify further: Simplify the equation further: $z = \frac{-6c \pm \sqrt{36c^2 - 8c^2}}{2c^2}$ .
Combine like terms: Combine like terms under the square root: $z = \frac{-6c \pm \sqrt{28c^2}}{2c^2}$ .
Simplify square root: Simplify the square root: $\sqrt{28c^2} = \sqrt{4\times7\times c^2} = 2c\times\sqrt{7}$ .
Cancel out terms: Now we have $z = \frac{-6c \pm 2c\sqrt{7}}{2c^2}$ . We can simplify this by canceling out a $2c$ in the numerator and denominator.
Cancel out terms: Now we have $z = \frac{-6c \pm 2c\sqrt{7}}{2c^2}$ . We can simplify this by canceling out a $2c$ in the numerator and denominator.After canceling, we get $z = \frac{-3 \pm \sqrt{7}}{c}$ . This gives us two possible solutions for $z$ .