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Big Ideas Math Algebra 2, 2014

BI

Big Ideas Math Algebra 2, 2014 View details

4. Solving Radical Equations and Inequalities

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Exercise 50 Page 267

We will solve the given system of equations using the Substitution Method.

{x^{2} + y^{2} = 25 y = - \frac{3}{4} x + \frac{2 5}{4} (I) (II)

The

y -

variable is isolated in Equation (II). This allows us to substitute its value

- \frac{3}{4} x + \frac{2 5}{4}

for

y

in Equation (I).

{x^{2} + y^{2} = 25 y = - \frac{3}{4} x + \frac{2 5}{4} (I) (II)

$(I):$ $y = - \frac{3}{4} x + \frac{2 5}{4}$

{x^{2} + (- \frac{3}{4} x + \frac{2 5}{4})^{2} = 25 y = - \frac{3}{4} x + \frac{2 5}{4} (I) (II)

▼

(I):

Simplify

ExpandPosPerfectSquare

$(I):$ $(a + b)^{2} = a^{2} + 2 a b + b^{2}$

{x^{2} + (- \frac{3}{4} x)^{2} + 2 (- \frac{3}{4} x) (\frac{2 5}{4}) + (\frac{2 5}{4})^{2} y = - \frac{3}{4} x + \frac{2 5}{4} (I) (II)

Calculate power and product

{x^{2} + \frac{9}{1 6} x - \frac{7 5}{8} x + \frac{6 2 5}{1 6} = 25 y = - \frac{3}{4} x + \frac{2 5}{4} (I) (II)

$(I):$ $LHS \cdot 16 = RHS \cdot 16$

{16 x^{2} + 9 x - 150 x + 625 = 400 y = - \frac{3}{4} x + \frac{2 5}{4} (I) (II)

$(I):$ Add terms

{25 x^{2} - 150 x + 625 = 400 y = - \frac{3}{4} x + \frac{2 5}{4} (I) (II)

$(I):$ $LHS - 400 = RHS - 400$

{25 x^{2} - 150 x + 225 = 0 y = - \frac{3}{4} x + \frac{2 5}{4} (I) (II)

$(I):$ $LHS / 25 = RHS / 25$

{x^{2} - 6 x + 9 = 0 y = - \frac{3}{4} x + \frac{2 5}{4} (I) (II)

Notice that in Equation (I) we have a quadratic equation in terms of only the

x -

variable. Notice that there are many ways to solve a quadratic equation. We will use Quadratic Formula. Let's determine

a,

b,

and

c .

x^{2} - 6 x + 9 = 0 \Leftrightarrow 1 x^{2} + (- 6) x + 9 = 0

We see above that

a = 1,

b = - 6,

and

c = 9 .

Let's substitute these values into the Quadratic Formula to solve the equation.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValues

Substitute values

x = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( 1 ) ( 9 )}{2 ( 1 )}

▼

Simplify right-hand side

Calculate power

x = \frac{- ( - 6 ) \pm 3 6 - 4 ( 1 ) ( 9 )}{2 ( 1 )}

Multiply

x = \frac{6 \pm 3 6 - 3 6}{2}

Subtract terms

x = \frac{6 \pm 0}{2}

Calculate root

x = \frac{6 \pm 0}{2}

Simplify numerator

x = \frac{6}{2}

Calculate quotient

x = 3

Now consider Equation (II) in the given nonlinear system.

y = - \frac{3}{4} x + \frac{2 5}{4}

We will substitute

x = 3

into the above equation to find the value for

y .

y = - \frac{3}{4} x + \frac{2 5}{4}

$x = 3$

y = - \frac{3}{4} (3) + \frac{2 5}{4}

▼

Simplify

Multiply

y = - \frac{9}{4} + \frac{2 5}{4}

Add fractions

y = \frac{1 6}{4}

Calculate quotient

y = 4

We found that

y = 4

when

x = 3 .

Therefore, the solution of the system is

(3, 4) .

Graphing

Let's notice that Equation (I) represents a circle and Equation (II) represents a line. If you need explanations on graphing circles, please refer to this site. If you need explanations on graphing lines, please refer to this site. Let's graph them and mark the point $(3, 4) .$

Notice that the circle and the line intersect about at the point $(3, 4),$ which confirms our answer.

Subchapter links

Communicate Your Answer p.261

Monitoring Progress p.262-265

Exercises p.266-268