Solve for x:

a)${(x+1)}^{2}=16$

b)${(3x-1)}^{2}=0$

c)${(x+8)}^{2}=-9$

a)

b)

c)

Anne Wacker
2021-12-14
Answered

Solve for x:

a)${(x+1)}^{2}=16$

b)${(3x-1)}^{2}=0$

c)${(x+8)}^{2}=-9$

a)

b)

c)

You can still ask an expert for help

lovagwb

Answered 2021-12-15
Author has **50** answers

Step 1

a)${(x+1)}^{2}=16$

$\Rightarrow (x+1)=\pm 4$

$\therefore x=3,\text{}-5$

$\therefore$ No of solutions is 2

Step 2

b)${(3x-1)}^{2}=0$

$\therefore x=\frac{1}{3}$

No of solutions$=1$

c)${(x+8)}^{2}=-9$

$\Rightarrow x+8-\sqrt{-9}$

$\Rightarrow x=\pm 3i-8$

$x=3i-8,\text{}-3i-8$

Here no real solution exist for x.

a)

Step 2

b)

No of solutions

c)

Here no real solution exist for x.

raefx88y

Answered 2021-12-16
Author has **26** answers

Step 1

a) Given:${(x+1)}^{2}=16$

Take the square root of both sides of the equation.

$\sqrt{{(x+1)}^{2}}=\sqrt{16}$

Simplify.

$x+1=4$

$x+1=-4$

Subtract 1 from both sides of the equation.

$x=3$

$x=-5$

Step 2

b) Given:${(3x-1)}^{2}=0$

Use binomial theorem$(a-b)}^{2}={a}^{2}-2ab+{b}^{2$ to expand $(3x-1)}^{2$

$9{x}^{2}-6x+1=0$

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 $\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$

$x=\frac{-(-6)\pm \sqrt{{(-6)}^{2}-4\times 9}}{2\times 9}$

Square -6.

$x=\frac{-(-6)\pm \sqrt{36-4\times 9}}{2\times 9}$

Multiply -4 times 9.

$x=\frac{-(-6)\pm \sqrt{36-36}}{2\times 9}$

Add 36 to -36.

$x=\frac{-(-6)\pm \sqrt{0}}{2\times 9}$

Take the square root of 0.

$x=\frac{-6}{2\times 9}$

The opposite of -6 is 6.

$x=\frac{6}{2\times 9}$

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:${(x+8)}^{2}=-9$

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.

$x=-8+3i$

$x=-8-3i$

a) Given:

Take the square root of both sides of the equation.

Simplify.

Subtract 1 from both sides of the equation.

Step 2

b) Given:

Use binomial theorem

This equation is in standard form:

Square -6.

Multiply -4 times 9.

Add 36 to -36.

Take the square root of 0.

The opposite of -6 is 6.

Multiply 2 times 9.

Reduce the fraction

Step 3

c) Given:

Subtract 8 from both sides of the equation.

Subtracting 8 from itself leaves 0.

Subtract 8 from 3i.

Subtract 8 from -3i.

The equation is now solved.

asked 2021-12-02

The function, $f\left(x\right)=-2{x}^{2}+x+5$ , is in standard form. The quadratic equation is $0=-2{x}^{2}+x+5$ , where $a=-2,\text{}b=1,$ and $c=5$ . The discriminate ${b}^{2}-4ac$ is 41. Now, complete step 5 to solve for the zeros of the quadratic function.

Solve using the quadratic formula.

$x=\frac{-b\pm \sqrt{{b}^{2}-4ax}}{2a}$

What are the zeros of the function

$f\left(x\right)=x+5-2{x}^{2}?$

a)$x=\frac{-1\pm \sqrt{41}}{-4}$

b)$x=\frac{1\pm \sqrt{41}}{-4}$

c)$x=\frac{-1\pm \sqrt{39}}{-4}$

d)$x=\frac{1\pm \sqrt{39}}{-4}$

Solve using the quadratic formula.

What are the zeros of the function

a)

b)

c)

d)

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Factoring Polynomial using Quadratic Equation

I'm trying to use the quadratic equation (QE) to factor a degree 2 polynomial into the format: $(x+a{)}^{2}$, where a is any real number.

This works great for equations like:

(1) ${x}^{2}+2x+1$

The QE gives roots $x=\{-1,-1\}$, and $(x+1{)}^{2}={x}^{2}+2x+1$.

This approach fails for other polynomials such as:

(2) $9{x}^{2}+36x+36$

The roots found with the QE are $x=\{-2,-2\}$. But $(x+2{)}^{2}\ne 9x+36x+36$.

9 is the GCF from the coefficients of (2), so (2) can be rewritten as:

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and by using the QE on the second term, this equals:

$9\times (x+2{)}^{2}$. And $9\times (x+2{)}^{2}=9x+36x+36$ (2), so this seems to be a better approach.

So my question is: must the input to QF be a polynomial that has coefficients with $GCF=1$, for the result to give roots that can be used to reconstruct the original function?

I'm trying to use the quadratic equation (QE) to factor a degree 2 polynomial into the format: $(x+a{)}^{2}$, where a is any real number.

This works great for equations like:

(1) ${x}^{2}+2x+1$

The QE gives roots $x=\{-1,-1\}$, and $(x+1{)}^{2}={x}^{2}+2x+1$.

This approach fails for other polynomials such as:

(2) $9{x}^{2}+36x+36$

The roots found with the QE are $x=\{-2,-2\}$. But $(x+2{)}^{2}\ne 9x+36x+36$.

9 is the GCF from the coefficients of (2), so (2) can be rewritten as:

$9\times ({x}^{2}+4x+4)$

and by using the QE on the second term, this equals:

$9\times (x+2{)}^{2}$. And $9\times (x+2{)}^{2}=9x+36x+36$ (2), so this seems to be a better approach.

So my question is: must the input to QF be a polynomial that has coefficients with $GCF=1$, for the result to give roots that can be used to reconstruct the original function?

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