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3 years ago

Can someone please explain in detail how to solve radical equations? (1, 2, &10)

Mathematics

1 answer:

BartSMP [9]3 years ago

8 0

Step-by-step explanation:

- First, the radical needs to be alone on one side of the equation, if there is - 4, - 3 next to it then you'll have to transfer it to the opposite side like that:

5. (y+7)^0.5 - 4=4

(y+7)^0.5=8......

- Second, you need to square the two sides of the equation :

5.((y+7)^0.5) ^2=8^2

Y+7=64

Y+7-7=64-7

Y=57

Note: if the radical's index is (2) then you'll square the two sides of the equation, but if the index is (3) you'll have to cube the two sides of the equation, "look at the exercises 9,10"

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3 years ago

For a triangle $XYZ$, we use $[XYZ]$ to denote its area. Let $ABCD$ be a square with side length $1$. Points $E$ and $F$ lie on

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An algebraic equation enables the expression of equality between variable expressions

$\underline{The \ value \ of \ [AEF] \ is \ \dfrac{4}{9}}$

The reason the above value is correct is given as follows:

The given parameters are;

The symbol for the area of a triangle ΔXYZ = [XYZ]

The side length of the given square ABCD = 1

The location of point E = Side $\overline{BC}$ on square ABCD

The location of point F = Side $\overline{CD}$ on square ABCD

∠EAF = 45°

The area of ΔCEF, [CEF] = 1/9 (corrected by using a similar online question)

Required:

To find the value of [AEF]

Solution:

The area of a triangle = (1/2) × Base length × Height

Let x = EC, represent the base length of ΔCEF, and let y = CF represent the height of triangle ΔCEF

We get;

The area of a triangle ΔCEF, [CEF] = (1/2)·x·y = x·y/2

The area of ΔCEF, [CEF] = 1/9 (given)

∴ x·y/2 = 1/9

ΔABE:

$\overline{BE}$ = BC - EC = 1 - x

The area of ΔABE, [ABE] = (1/2)×AB ×BE

AB = 1 = The length of the side of the square

The area of ΔABE, [ABE] = (1/2)× 1 × (1 - x) = (1 - x)/2

ΔADF:

$\overline{DF}$ = CD - CF = 1 - y

The area of ΔADF, [ADF] = (1/2)×AD ×DF

AD = 1 = The length of the side of the square

The area of ΔADF, [ADF] = (1/2)× 1 × (1 - y) = (1 - y)/2

The area of ΔAEF, [AEF] = [ABCD] - [ADF] - [ABE] - [CEF]

[ABCD] = Area of the square = 1 × 1

$[AEF] = 1 - \dfrac{1 - x}{2} - \dfrac{1 - y}{2} - \dfrac{1}{19}= \dfrac{19 \cdot x + 19 \cdot y - 2}{38}$

From $\dfrac{x \cdot y}{2} = \dfrac{1}{9}$ , we have;

$x = \dfrac{2}{9 \cdot y}$ , which gives;

$[AEF] = \dfrac{9 \cdot x + 9 \cdot y - 2}{18}$

Area of a triangle = (1/2) × The product of the length of two sides × sin(included angle between the sides)

∴ [AEF] = (1/2) × $\overline{AE}$ × $\overline{FA}$ × sin(∠EAF)

$\overline{AE}$ = √((1 - x)² + 1), $\overline{FA}$ = √((1 - y)² + 1)

[AEF] = (1/2) × √((1 - x)² + 1) × √((1 - y)² + 1) × sin(45°)

Which by using a graphing calculator, gives;

$\dfrac{1}{2} \times \sqrt{(1 - x)^2 + 1} \times \sqrt{(1 - y)^2 + 1} \times \dfrac{\sqrt{2} }{2} = \dfrac{9 \cdot x + 9 \cdot y - 2}{18}$

Squaring both sides and plugging in $x = \dfrac{2}{9 \cdot y}$ , gives;

$\dfrac{(81 \cdot y^4-180 \cdot y^3 + 200 \cdot y^2 - 40\cdot y +4)\cdot y^2}{324\cdot y^4} = \dfrac{(81\cdot y^4-36\cdot y^3 + 40\cdot y^2 - 8\cdot y +4)\cdot y^2}{324\cdot y^2}$

Subtracting the right hand side from the equation from the left hand side gives;

$\dfrac{40\cdot y- 36\cdot y^2 + 8}{81\cdot y} = 0$

36·y² - 40·y + 8 = 0

$y = \dfrac{40 \pm \sqrt{(-40)^2-4 \times 36\times 8} }{2 \times 36} = \dfrac{5 \pm \sqrt{7} }{9}$

$[AEF] = \dfrac{9 \cdot x + 9 \cdot y - 2}{18} = \dfrac{9 \cdot y^2-2 \cdot y + 2}{18 \cdot y}$

Plugging in $y = \dfrac{5 + \sqrt{7} }{9}$ and rationalizing surds gives;

$[AEF] = \dfrac{9 \cdot \left(\dfrac{5 + \sqrt{7} }{9}\right) ^2-2 \cdot \left(\dfrac{5 + \sqrt{7} }{9}\right) + 2}{18 \cdot \left(\dfrac{5 + \sqrt{7} }{9}\right) } = \dfrac{\dfrac{40+8\cdot \sqrt{7} }{9} }{10+2\cdot \sqrt{7} } = \dfrac{32}{72} = \dfrac{4}{9}$