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SAT II Math I : Solving Equations

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Example Questions

Example Question #253 : Sat Subject Test In Math I

Give the set of all real solutions of the following equation:

$\frac{8}{x^{2}+4x+4 } - \frac{2}{x+ 2 } - 3 = 0$

Possible Answers:

$\left \{ - \frac{2}{3} \right \}$

$\left \{ \frac{2}{3} \right \}$

$\left \{ -4, - \frac{2}{3} \right \}$

No solution

$\left \{ 4, \frac{2}{3} \right \}$

Correct answer:

$\left \{ -4, - \frac{2}{3} \right \}$

Explanation:

$x^{2} + 4x + 4$ can be seen to fit the perfect square trinomial pattern:

$x^{2} + 4x + 4 = x^{2} + 2 \cdot x \cdot 2 + 2 ^{2 } = (x + 2) ^{2 }$

The equation can therefore be rewritten as

$\frac{8}{ (x + 2) ^{2 }} - \frac{2}{x+ 2 } - 3 = 0$

Multiply both sides of the equation by the least common denominator of the expressions, which is $LCM (x+2 , (x+2) ^{2}) = (x+2)^{2}$ :

$\left [\frac{8}{ (x + 2) ^{2 }} - \frac{2}{x+ 2 } - 3 \right ] (x+2)^{2} = 0 \cdot (x+2)^{2}$

$\frac{8}{ (x + 2) ^{2 }}\cdot (x+2)^{2} - \frac{2}{x+ 2 }\cdot (x+2)^{2} - 3 \cdot (x+2)^{2} = 0$

$8 - 2 (x+2) - 3 \cdot (x+2)^{2} = 0$

Distributing and collecting like terms, we get

$8 - 2 (x+2) - 3 \cdot (x^{2} + 4x + 4 ) = 0$

$8 - 2 \cdot x - 2 \cdot 2 - 3 \cdot x^{2} - 3\cdot 4x - 3\cdot 4 = 0$

$8 - 2 x - 4 - 3 x^{2} -12 x - 12= 0$

$- 3 x^{2} - 2 x -12 x + 8 - 4 - 12= 0$

$- 3 x^{2} - 14 x -8= 0$

Multiplying both sides by :

$\left (- 3 x^{2} - 14 x -8 \right )(-1)= 0 (-1)$

$3 x^{2} + 14 x +8 = 0$

This can be solved using the method. We are looking for two integers whose sum is and whose product is $3 \times 8 = 24$ . Through some trial and error, the integers are found to be 2 and 12, so the above equation can be rewritten, and solved using grouping, as

$3 x^{2} + 12x + 2x +8 = 0$

$(3 x^{2} + 12x) + (2x +8 )= 0$

3x (x+4 ) + 2 (x+4 )= 0

(3x + 2 )(x+4 )= 0

By the Zero Product Principle, one of these factors is equal to zero:

Either:

3x + 2 = 0

3x + 2 - 2 = 0 - 2

3x = -2

$\frac{3x }{3}= \frac{-2}{3}$

$x = -\frac{2}{3}$

Or:

x+ 4 = 0

x+ 4 - 4= 0 - 4

x = -4

Both solutions can be confirmed by substitution; the solution set is $\left \{ -4, - \frac{2}{3} \right \}$

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Example Question #91 : Solving Equations

A polynomial of degree four has as its lead term $x^{4}$ and has rational coefficients. Two of its zeroes are 2-3i and 3+ 2 i . What is this polynomial?

Possible Answers:

$x^{4} + 10x^{3} + 50x^{2} + 130x +169$

$x^{4} - 10x^{3} + 50x^{2} - 130x +169$

$x^{4} -13x^{3} + 26x^{2} + 65x +25$

Cannot be determined

$x^{4} +13x^{3} + 26x^{2} - 65x +25$

Correct answer:

$x^{4} - 10x^{3} + 50x^{2} - 130x +169$

Explanation:

A fourth-degree, or quartic, polynomial has four zeroes, if a zero of multiplicity is counted times. Since its lead term is $x^{4}$ , we know that

$p(x) = (x-b_{1}) (x-b_{2}) (x-b_{3})(x-b_{4})$

A polynomial with rational coefficients has its imaginary zeroes in conjugate pairs. Since p(x) is such a polynomial, then, since 2-3i is a zero, so is its complex conjugate 2+3i ; similarly, since 3+ 2 i is a zero, so is its complex conjugate 3- 2 i . Substituting these four values for the four $b_{n}$ values:

p(x) = [x-(2+3i)][x-(2-3i)] [x-(3+2i)] [x-(3-2i)]

This can rewritten as

p(x) =[x- 2+3i][x- 2-3i] [x- 3+2i][x- 3-2i]

p(x) =[(x- 2)+3i][(x- 2)-3i] [(x- 3)+2i][(x- 3)-2i]

Multiply the first two factors using the difference of squares pattern, then the square of a binomial pattern:

[(x- 2)+3i][(x- 2)-3i]

$= (x- 2)^{2}-(3i)^{2}$

$= (x- 2)^{2}-9i^{2}$

$= x^{2} -4x +4 +9$

$= x^{2} -4x +13$

Multiply the last two factors similarly:

[(x- 3)+2i][(x- 3)-2i]

$= (x- 3)^{2}-(2i)^{2}$

$= (x- 3)^{2}-4i^{2}$

$= x^{2} -6x+9 +4$

$= x^{2} -6x +13$

Thus,

$p(x) = ( x^{2} -4x +13) ( x^{2} -6x +13)$

Multiply:

$x^{2} -4x +13$

$\underline{ x^{2} -6x +13}$

$13x^{2} -52x +169$

$-6 x^{3} +24x^{2} -78x$

$\underline{x^{4} -4x^{3} +13x^{2}}$ ________________

$x^{4} - 10x^{3} + 50x^{2} - 130x +169$

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Example Question #101 : Single Variable Algebra

Give the set of all real solutions of the equation $8 x -2 x ^{\frac{1}{2}} - 3 = 0$ .

Possible Answers:

$\left \{ - \frac{1}{4} , \frac{1}{4} \right \}$

$\left \{ \frac{1}{4} , \frac{9}{16} \right \}$

None of these

$\left \{ - \frac{9}{16} , \frac{9}{16} \right \}$

$\left \{ - \frac{1}{4} , \frac{9}{16} \right \}$

Correct answer:

None of these

Explanation:

Set $y = x^{\frac{1}{2}}$ . Then $x = \left (x^{\frac{1}{2}} \right )^{2} = y ^{2}$ .

$8 x -2 x ^{\frac{1}{2}} - 3 = 0$ can be rewritten as

$8 \left (x^{\frac{1}{2}} \right )^{2} -2 x ^{\frac{1}{2}} - 3 = 0$

Substituting $y^{2}$ for and for $x^{\frac{1}{2}}$ , the equation becomes

$8y^{2} -2 y - 3 = 0$ ,

a quadratic equation in .

This can be solved using the method. We are looking for two integers whose sum is and whose product is 8 (-3) = -24 . Through some trial and error, the integers are found to be and 4, so the above equation can be rewritten, and solved using grouping, as

$8y^{2}-6y + 4 y - 3 = 0$

2y ( 4 y - 3)+1 ( 4 y - 3) = 0

(2y +1) ( 4 y - 3) = 0

By the Zero Product Principle, one of these factors is equal to zero:

Either:

2y + 1 = 0

2y + 1 - 1 = 0 - 1

2y = - 1

$\frac{2y}{2} = \frac{- 1}{2}$

$y = - \frac{1}{2}$

Substituting back:

$x^{\frac{1}{2}} = -\frac{1}{2}$ , or $\sqrt{x} = - \frac{1}{2}$

However, this does not hold for any real value of . No solution is yielded here.

Or:

4y - 3 = 0

4y - 3+ 3 = 0+ 3

4 y = 3

$\frac{4 y}{4} = \frac{3}{4}$

$y= \frac{3}{4}$

Substituting back:

$x^{\frac{1}{2}} = \frac{3}{4}$ , or $\sqrt{x} = \frac{3}{4}$

$(\sqrt{x} )^{2} = \left (\frac{3}{4} \right )^{2}$

$x = \frac{9}{16}$ ,

the only solution. This is not among the choices.

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SAT II Math I : Solving Equations

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #253 : Sat Subject Test In Math I

Example Question #91 : Solving Equations

Example Question #101 : Single Variable Algebra

All SAT II Math I Resources

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