GED Math - Solving by Other Methods

What is the solution to the equation ? Round your answer to the nearest tenths place.

Explanation

Recall the quadratic equation:

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

For the given equation, . Plug these into the equation and solve.

$x=\frac{-5+\sqrt{5^2-4(1)(-7))}}{2(1)}=\frac{-5+\sqrt{25+28}}{2}=\frac{-5+\sqrt{53}}{2}=1.1$

and

$x=\frac{-5-\sqrt{5^2-4(1)(-7))}}{2(1)}=\frac{-5-\sqrt{25+28}}{2}=\frac{-5-\sqrt{53}}{2}=-6.1$

Solve the following for x by completing the square:

$x^{2}+8x-3=0$

$x=-4+\sqrt{19}$ or $x=-4-\sqrt{19}$

$x=4+\sqrt{19}$ or $x=4-\sqrt{19}$

$x=-4+\sqrt{19}$

$x=-4+\sqrt{3}$ or $x=-4-\sqrt{3}$

$x=-4+\sqrt{67}$ or $x=-4-\sqrt{67}$

Explanation

To complete the square, we need to get our variable terms on one side and our constant terms on the other.

$x^{2}+8x-3=0$

$x^{2}+8x=3$

To make a perfect square trinomial, we need to take one-half of the x-term and square said term. Add the squared term to both sides.

$x^{2}+8x+{\color{Blue} 16}=3+{\color{Blue} 16}$

$x^{2}+8x+16=19$

We now have a perfect square trinomial on the left side which can be represented as a binomial squared. We should check to make sure.

* $a^{2}+2ab+b^{2}=(a+b)^{2}$ (standard form)

In our equation:

$(x)^{2}+8x+(4)^{2}$

(CHECK)

Represent the perfect square trinomial as a binomial squared:

$x^{2}+4x+16=(x+4)^{2}$

Take the square root of both sides:

$(x+4)^{2}=19$

$\sqrt{(x+4)^{2}}=_{-}^{+}\sqrt{19}$

$(x+4)=_{-}^{+}\sqrt{19}$

Solve for x

$x= -4_{-}^{+}\sqrt{19}$

$x=-4+\sqrt{19}$ or $x=-4-\sqrt{19}$

Solve the following by using the Quadratic Formula:

$x^{2}+ 4=0$

$x=_{-}^{+}\textrm{2}i$

No solution

Explanation

The Quadratic Formula:

$x=\frac{-b_{-}^{+}\sqrt{b^{2}-4ac}}{2a}$

Plugging into the Quadratic Formula, we get

$x=\frac{-0_{-}^{+}\sqrt{(0)^{2}-4(1)(4)}}{2(1)}$

$x=\frac{_{-}^{+}\sqrt{0-16}}{2}$

$x=\frac{_{-}^{+}\sqrt{-16}}{2}$

*The square root of a negative number will involve the use of complex numbers

$\sqrt{-16}=\sqrt{(16)(-1)}=(\sqrt{16})(\sqrt{-1})=4(\sqrt{-1})$

$\sqrt{-1}=i$

Therefore, $\sqrt{-16}=4i$

$x=\frac{_{-}^{+}4i}{2}$

$x=_{-}^{+}2i$

Solve for :

$4x ^{2} - 20x + 25 = 0$

$x = 2\frac{1}{2}$

$x = \frac{2}{5}$

$x = - 2\frac{1}{2} \textrm{ or }x = 2\frac{1}{2}$

$x = -\frac{2}{5} \textrm{ or }x = \frac{2}{5}$

Explanation

$4x ^{2} - 20x + 25$ can be demonstrated to be a perfect square polynomial as follows:

$4x ^{2} - 20x + 25 = \left ( 2x\right )^{2} - 2 (2x)(5) + 5^{2}$

It can therefore be factored using the pattern

$A^{2}-2AB + B^{2} = \left (A - B \right )^{2}$

with .

We can rewrite and solve the equation accordingly:

$4x ^{2} - 20x + 25 = 0$

$(2x-5) ^{2} = 0$

$x = 5 \div 2$

$x = 2\frac{1}{2}$

This is the only solution.

Solve for :

$x^{2} - 12 = 6x+4$

Explanation

When solving a quadratic equation, it is necessary to write it in standard form first - that is, in the form $ax^{2} + bx + c = 0$ . This equation is not in this form, so we must get it in this form as follows:

$x^{2} - 12 = 6x+4$

$x^{2} - 12- 6x - 4 = 6x+4 - 6x - 4$

$x^{2} - 6x - 16 = 0$

We factor the quadratic expression as

so that and .

By trial and error, we find that

, so the equation becomes

Set each linear binomial to 0 and solve separately:

The solutions set is $\left { -2, 8\right }$

Solve for x by using the Quadratic Formula:

$2x^{2}+7x-85=0$

x = 5 or x= -8.5

x = 5

x = 10 or x = -17

x = -8.5

x = -5 or x = 8.5

Explanation

We have our quadratic equation in the form $ax^{2}+bx+c=0$

The quadratic formula is given as:

$x=\frac{-b_{-}^{+}{\sqrt{b^{2}-4ac}}}{2a}$

Using $2x^{2}+7x-85=0$

$x=\frac{-7_{-}^{+}{\sqrt{(7)^{2}-4(2)(-85)}}}{2(2)}$

$x=\frac{-7_{-}^{+}{\sqrt{49+680}}}{4}$

$x=\frac{-7_{-}^{+}{\sqrt{729}}}{4}$

$x=\frac{-7_{-}^{+}{27}}{4}$

$x=\frac{-7+27}{4}$ $x=\frac{-7-27}{4}$

$x=\frac{20}{4}$ $x=\frac{-34}{4}$

$\textbf{x=5}$ $\textbf{x=-8.5}$

Solve for :

Explanation

$x \cdot x -x \cdot 7 = 30$

$x ^{2} -7x = 30$

$x ^{2} -7x - 30 = 0$

We factor the quadratic expression as

so that and .

By trial and error, we find that

, so the equation becomes

Set each linear binomial to 0 and solve separately:

The solution set is $\left { -3, 10 \right }$ .

What is the solution to the equation ? Round your answer to the nearest hundredths place.

Explanation

Solve this equation by using the quadratic equation:

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

For the equation ,

Plug it in to the equation to solve for .

$x=\frac{-4\pm \sqrt{4^2-4(1)(-10)}}{2}$

$x=\frac{-4\pm \sqrt{16+40}}{2}$

$x=\frac{-4\pm \sqrt{56}}{2}$

and

A rectangular yard has a width of w and a length two more than three times the width. The area of the yard is 120 square feet. Find the length of the yard.

20 feet

6 feet

89 feet

24 feet

5 feet

Explanation

The area of the garden is 120 square feet. The width is given by w, and the length is 2 more than 3 times the width. Going by the order of operations implied, we have length given by 3w+2.

(length) x (width) = area (for a rectangle)