# Solve for x: a) (x+1)^{2}=16 b) (3x-1)^{2}=0 c) (x+8)^{2}=-9

Solve for x:
a) ${\left(x+1\right)}^{2}=16$
b) ${\left(3x-1\right)}^{2}=0$
c) ${\left(x+8\right)}^{2}=-9$
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lovagwb
Step 1
a) ${\left(x+1\right)}^{2}=16$
$⇒\left(x+1\right)=±4$

$\therefore$ No of solutions is 2
Step 2
b) ${\left(3x-1\right)}^{2}=0$
$\therefore x=\frac{1}{3}$
No of solutions $=1$
c) ${\left(x+8\right)}^{2}=-9$
$⇒x+8-\sqrt{-9}$
$⇒x=±3i-8$

Here no real solution exist for x.
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raefx88y
Step 1
a) Given: ${\left(x+1\right)}^{2}=16$
Take the square root of both sides of the equation.
$\sqrt{{\left(x+1\right)}^{2}}=\sqrt{16}$
Simplify.
$x+1=4$
$x+1=-4$
Subtract 1 from both sides of the equation.
$x=3$
$x=-5$
Step 2
b) Given: ${\left(3x-1\right)}^{2}=0$
Use binomial theorem ${\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}$ to expand ${\left(3x-1\right)}^{2}$
$9{x}^{2}-6x+1=0$
This equation is in standard form: $a{x}^{2}+bx+c=0$ NSk Substitute 9 for a, -6 for b, and 1 for c in the quadratic formula $\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
$x=\frac{-\left(-6\right)±\sqrt{{\left(-6\right)}^{2}-4×9}}{2×9}$
Square -6.
$x=\frac{-\left(-6\right)±\sqrt{36-4×9}}{2×9}$
Multiply -4 times 9.
$x=\frac{-\left(-6\right)±\sqrt{36-36}}{2×9}$
$x=\frac{-\left(-6\right)±\sqrt{0}}{2×9}$
Take the square root of 0.
$x=\frac{-6}{2×9}$
The opposite of -6 is 6.
$x=\frac{6}{2×9}$
Multiply 2 times 9.
$x=\frac{6}{18}$
Reduce the fraction $\frac{6}{18}$ to lowest terms by extracting and canceling out 6.
$x=\frac{1}{3}$
Step 3
c) Given: ${\left(x+8\right)}^{2}=-9$
Subtract 8 from both sides of the equation.
$x+8-8=3i-8$
$x+8-8=-3i-8$
Subtracting 8 from itself leaves 0.
$x=3i-8$
$x=-3i-8$
Subtract 8 from 3i.
$x=-8+3i$
Subtract 8 from -3i.
$x=-8-3i$
The equation is now solved.
$x=-8+3i$
$x=-8-3i$