Understand and apply concepts of equations - HiSET

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Question

Solve.

$\frac{x}{4}+11=15$

Answer

In order to solve for the variable, , we need to isolate it on the left side of the equation. We will do this by reversing the operations done to the variable by performing the opposite of each operation on both sides of the equation.

Let's begin by rewriting the given equation.

$\frac{x}{4}+11=15$

Subtract from both sides of the equation.

$\frac{x}{4}+11-11=15-11$

Simplify.

$\frac{x}{4}=4$

Multiply both sides of the equation by .

$\frac{x}{4}\times 4=4\times 4$

Solve.

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Question

Give the solution set of the inequality

$\left | -4x + 13 \right | \ge 9$

Answer

$\left | -4x + 13 \right | \ge 9$ can be rewritten as the compound inequality

$-4x + 13 \le - 9 \textup{ or } -4x + 13 \ge 9$ .

The solution set will be the union of the two individual solution sets. Find the solution set of the first inequality as follows:

Isolate by first subtracting from both sides:

$-4x + 13 \le - 9$

$-4x + 13- 13 \le - 9 - 13$

$-4x \le -22$

Divide both sides by , reversing the direction of the inequality symbol, since you are dividing by a negative number:

$-4x \div (-4){ \color{Red} \ge} -22 \div (-4)$

$x \ge 5\frac{1}{2}$ .

In interval notation, this is the set $\left [ 5 \frac{1}{2} , \infty \right )$ .

Find the solution set of the other inequality similarly:

$-4x + 13 \ge 9$

$-4x + 13 - 13 \ge 9 - 13$

$-4x \ge -4$

$-4x \div (-4) \le -4 \div (-4)$

$x \le 1$

In interval notation, this is the set $\left ( -\infty, 1 \right ]$ .

The union of these sets is the solution set: $\left ( -\infty, 1 \right ] \cup \left [ 5 \frac{1}{2} , \infty \right )$ .

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Question

What is the vertex of the following quadratic polynomial?

Answer

Given a quadratic function

the vertex will always be

$(-\frac{b}{2a},\frac{4ac-b^2}{4a})$ .

Thus, since our function is

, , and .

We plug these variables into the formula to get the vertex as

$(-\frac{-3}{2\times2},\frac{4\times2\times17-(-3)^2}{4\times2})$

$(-\frac{-3}{4},\frac{136-9}{8})$

$=(\frac{3}{4},\frac{127}{8})$ .

Hence, the vertex of

is

$(\frac{3}{4},\frac{127}{8})$ .

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Question

Which of the following expressions represents the discriminant of the following polynomial?

Answer

The discriminant of a quadratic polynomial

is given by

.

Thus, since our quadratic polynomial is

,

, , and .

Plugging these values into the discriminant equation, we find that the discriminant is

$9-4\times6\times7$ .

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Question

Give the nature of the solution set of the equation

$2x^{2}+ 5x = -17$

Answer

To determine the nature of the solution set of a quadratic equation, it is necessary to first express it in standard form

$ax^{2}+bx+c = 0$

This can be done by adding 17 to both sides:

$2x^{2}+ 5x+ 17 = -17+ 17$

$2x^{2}+ 5x+ 17 = 0$

The key to determining the nature of the solution set is to examine the discriminant

$b^{2} - 4ac$ . Setting , the value of the discriminant is

$5^{2} - 4(2)(17) = 25 - 136 = -111$

This value is negative. Consequently, the solution set comprises two imaginary numbers.

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Question

Which of the following polynomial equations has exactly one solution?

Answer

A polynomial equation of the form

$ax^{2}+bx+c = 0$

has one and only one (real) solution if and only if its discriminant is equal to zero - that is, if its coefficients satisfy the equation

$b^{2}-4ac = 0$

In each of the choices, and , so it suffices to determine the value of which satisfies this equation. Substituting, we get

$b^{2}-4 (4)(25) = 0$

$b^{2}-400 = 0$

Solve for by first adding 400 to both sides:

$b^{2}-400 + 400 = 0 + 400$

$b^{2} = 400$

Take the square root of both sides:

$b =\pm \sqrt{400} = \pm 20$

The choice that matches this value of is the equation

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

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Question

Give the nature of the solution set of the equation

Answer

To determine the nature of the solution set of a quadratic equation, it is necessary to first express it in standard form

$ax^{2}+bx+c = 0$

To accomplish this, first, multiply the binomials on the left using the FOIL technique:

$x^{2}-5x -4x+20= 18$

Collect like terms:

$x^{2}-9x+20= 18$

Now, subtract 18 from both sides:

$x^{2}-9x+20 -18 = 18 -18$

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

The key to determining the nature of the solution set is to examine the discriminant $b^{2} - 4ac$ . Setting , the value of the discriminant is

$b^{2} - 4ac = (-9)^{2} - 4(1)(2)= 81 - 8 = 73$

The discriminant is a positive number, so there are two real solutions. Since 73 is not a perfect square, the solutions are irrational.

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Question

Give the nature of the solution set of the equation

Answer

To determine the nature of the solution set of a quadratic equation, it is necessary to first express it in standard form

$ax^{2}+bx+c = 0$

To accomplish this, first, multiply the binomials on the left using the FOIL technique:

$2x ^{2}+8x -5x-20= -6$

Collect like terms:

$2x ^{2}+3x-20= -6$

Add 6 to both sides:

$2x ^{2}+3x-20 + 6 = -6 + 6$

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

The key to determining the nature of the solution set is to examine the discriminant $b^{2} - 4ac$ . Setting , the value of the discriminant is

$b^{2} - 4ac= 3^{2} -4(2)(-14) = 9 - (-112) = 9+112 = 121$

The discriminant is a positive number; furthermore, it is a perfect square, being equal to the square of 11. Therefore, the solution set comprises two rational solutions.

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Question

Give the nature of the solution set of the equation

$6x+ 2x^{2}+ 7= 0$ .

Answer

To determine the nature of the solution set of a quadratic equation, it is necessary to first express it in standard form

$ax^{2}+bx+c = 0$

This can be done by simply switching the first and second terms:

$6x+ 2x^{2}+ 7= 0$

$2x^{2}+6x+ 7= 0$

The key to determining the nature of the solution set is to examine the discriminant $b^{2} - 4ac$ . Setting , the value of the discriminant is

$b^{2} - 4ac =6^{2} - 4(2)(7) = 36-56=-20$

The discriminant has a negative value. It follows that the solution set comprises two imaginary values.

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Question

Which of the following polynomial equations has exactly one solution?

Answer

A polynomial equation of the standard form

$ax^{2}+bx+c = 0$

has one and only one (real) solution if and only if its discriminant is equal to zero - that is, if its coefficients satisfy the equation

$b^{2}-4ac = 0$

Each of the choices can be rewritten in standard form by subtracting the term on the right from both sides. One of the choices can be rewritten as follows:

$9x^{2} + 25 =10x$

$9x^{2} + 25 - 10x =10x - 10x$

$9x^{2} - 10x + 25 =0$

By similar reasoning, the other four choices can be written:

$9x^{2} - 20x + 25 =0$

$9x^{2} - 30x + 25 =0$

$9x^{2} - 40x + 25 =0$

$9x^{2} - 50x + 25 =0$

In each of the five standard forms, and , so it is necessary to determine the value of that produces a zero discriminant. Substituting accordingly:

$b^{2}-4ac = 0$

$b^{2}-4(9)(25) = 0$

$b^{2}-900 = 0$

Add 900 to both sides and take the square root:

$b^{2}-900 + 900 = 0+ 900$

$b^{2}= 900$

$b = \pm \sqrt{900} = \pm 30$

Of the five standard forms,

$9x^{2} - 30x + 25 =0$

fits this condition. This is the standard form of the equation

$9x^{2} + 25 =30x$ ,

the correct choice.

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Question

Give the nature of the solution set of the equation

Answer

To determine the nature of the solution set of a quadratic equation, it is necessary to first express it in standard form

$ax^{2}+bx+c = 0$

To accomplish this, first, multiply the binomials on the left using the FOIL technique:

$x^{2}-5x -4x+20= -18$

Collect like terms:

$x^{2}-9x+20=- 18$

Now, add 18 to both sides:

$x^{2}-9x+20 +18 = -18+18$

$x^{2}-9x+38 = 0$

The key to determining the nature of the solution set is to examine the discriminant $b^{2} - 4ac$ . Setting , the value of the discriminant is

$b^{2} - 4ac = (-9)^{2} - 4(1)(38)= 81 - 152 = -71$

This discriminant is negative. Consequently, the solution set comprises two imaginary numbers.

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Question

Give the nature of the solution set of the equation

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

Answer

To determine the nature of the solution set of a quadratic equation, it is necessary to first express it in standard form

$ax^{2}+bx+c = 0$

This can be done by switching the first and third terms on the left:

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

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

The key to determining the nature of the solution set is to examine the discriminant

$b^{2} - 4ac$ . Setting , the value of the discriminant is

$4^{2} - 4 (-1)(20)= 16 - (-80 )= 96$ .

The discriminant is a positive number but not a perfect square. Therefore, there are two irrational solutions.

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Question

Give the nature of the solution set of the equation

$2x^{2}+ 5x =34$

Answer

To determine the nature of the solution set of a quadratic equation, it is necessary to first express it in standard form

$ax^{2}+bx+c = 0$

This can be done by subtracting 34 from to both sides:

$2x^{2}+ 5x =34$

$2x^{2}+ 5x- 34 =34 - 34$

$2x^{2}+ 5x- 34 =0$

The key to determining the nature of the solution set is to examine the discriminant

$b^{2} - 4ac$ . Setting , the value of the discriminant is

$5^{2} - 4(2)(-34) = 25 - (-272) = 297$ .

The discriminant is a positive number but not a perfect square. Therefore, there are two irrational solutions.

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Question

Solve for :

$\frac{2x+3y }{c} = 20$

Answer

To solve for in a literal equation, use the properties of algebra to isolate on one side, just as if you were solving a regular equation.

$\frac{2x+3y }{c} = 20$

Multiply both sides by :

$\frac{2x+3y }{c} \cdot c = 20 \cdot c$

Subtract from both sides:

Multiply both sides by $\frac{1}{2}$ , distributing on the right:

$\frac{1}{2} \cdot 2x =\frac{1}{2} \cdot (20 c - 3y)$

$x =\frac{1}{2} \cdot 20 c -\frac{1}{2} \cdot 3y$

$x =10 c -\frac{3}{2} y$ ,

the correct response.

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Question

Solve for :

$9x^{2}+z = y$

Assume is positive.

Answer

To solve for in a literal equation, use the properties of algebra to isolate on one side, just as if you were solving a regular equation.

$9x^{2}+z = y$

Subtract from both sides:

$9x^{2}+z - z = y - z$

$9x^{2} = y - z$

Divide both sides by 9:

$\frac{9x^{2}}{9} = \frac{y - z}{9}$

$x^{2} = \frac{y - z}{9}$

Take the square root of both sides:

$x =\sqrt{ \frac{y - z}{9}}$

Simplify the expression on the right by splitting it, and taking the square root of numerator and denominator:

$x = \frac{\sqrt {y - z}}{\sqrt {9}}$

$x = \frac{\sqrt {y - z}}{3}$ ,

the correct response.

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Question

Solve for :

Answer

To solve for in a literal equation, use the properties of algebra to isolate on one side, just as if you were solving a regular equation.

Subtract 20 from both sides:

Divide both sides by :

$\frac{xy}{y} = \frac{z - 20}{y}$

$x= \frac{z - 20}{y}$ ,

the correct response.

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Question

Solve for :

$\sqrt{x+1}= y+1$

Answer

To solve for in a literal equation, use the properties of algebra to isolate on one side, just as if you were solving a regular equation.

First, square both sides to eliminate the radical symbol:

$\sqrt{x+1}= y+1$

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

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

Rewrite the expression on the right using the square of a binomial pattern:

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

$x+1= y^{2}+ 2 y+1$

Subtract 1 from both sides:

$x+1-1 = y^{2}+ 2 y+1 - 1$

$x = y^{2}+ 2 y$ ,

the correct response.

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Question

Solve for :

$32 -x^{2} = y$

You my assume is positive.

Answer

To solve for in a literal equation, use the properties of algebra to isolate on one side, just as if you were solving a regular equation.

First, add $x^{2} - y$ to both sides:

$32 -x^{2} +(x^{2} - y) = y +(x^{2} - y)$

$32 -x^{2} +x^{2} - y = x^{2} +y - y$

$32 - y = x^{2}$

Take the positive square root of both sides:

$x=\sqrt{ 32- y}$ ,

the correct response.

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Question

Solve for :

$\frac{y+1}{x+1} = 3$

Answer

To solve for in a literal equation, use the properties of algebra to isolate on one side, just as if you were solving a regular equation.

First, take the reciprocal of both sides:

$\frac{y+1}{x+1} = 3$

$\frac{x+1}{y+1} = \frac{1}{3}$

Multiply both sides by :

$\frac{x+1}{y+1} \cdot (y+1)= \frac{1}{3}(y+1)$

$x+1= \frac{1}{3}(y+1)$

Distribute on the right:

$x+1= \frac{1}{3} \cdot y+ \frac{1}{3} \cdot 1$

$x+1= \frac{1}{3} y+ \frac{1}{3}$

Subtract 1 from both sides, rewriting 1 as $\frac{3}{3}$ to facilitate subtraction:

$x+1 - 1 = \frac{1}{3} y+ \frac{1}{3} - 1$

$x = \frac{1}{3} y+ \frac{1}{3} - \frac{3}{3}$

$x = \frac{1}{3} y - \frac{2}{3}$ ,

the correct response.

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Question

Solve.

$\frac{x}{4}+11=15$

Answer

In order to solve for the variable, , we need to isolate it on the left side of the equation. We will do this by reversing the operations done to the variable by performing the opposite of each operation on both sides of the equation.

Let's begin by rewriting the given equation.

$\frac{x}{4}+11=15$

Subtract from both sides of the equation.

$\frac{x}{4}+11-11=15-11$

Simplify.

$\frac{x}{4}=4$

Multiply both sides of the equation by .

$\frac{x}{4}\times 4=4\times 4$

Solve.

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