How to find the solution to an equation - ISEE Upper Level Quantitative Reasoning

First, rewrite the quadratic equation in standard form by distributing the through the product on the left and collecting all of the terms on the left side:

Use the method to factor the quadratic expression ; we are looking to split the linear term by finding two integers whose sum is and whose product is . These integers are , so:

Set each expression equal to 0 and solve:

or

The solution set is .

First, rewrite the quadratic equation in standard form by FOILing out the product on the left, then collecting all of the terms on the left side:

Use the method to split the middle term into two terms; we want the coefficients to have a sum of 1 and a product of . These numbers are , so we do the following:

Set each expression equal to 0 and solve:

or

The solution set is .

has as its domain the set of values of for which its radicand is nonnegative; that is,

or

Similarly, has as its domain the set of values of for which its radicand is nonnegative; that is,

or

The domain of the sum of two functions is the intersection of the domains of the two individual functions. This intersection is

The domain of is the single value , which is not among the choices.

Examine this diagram:

As indicated by the blue circle in the diagram, the only point on the graph of the function with domain element 4 is , so , and .

As indicated by the red circles in the diagram, there are three points on the graph of the function with range element 4 - namely, , , and . Therefore, . It is unclear whether or without further information.

First, rewrite the quadratic equation in standard form by distributing the through the product on the left, then collecting all of the terms on the left side:

Use the method to factor the quadratic expression ; we are looking to split the linear term by finding two integers whose sum is 7 and whose product is . These integers are , so:

Set each expression equal to 0 and solve:

or

The solution set is .

By the Zero Product Principle:

, in which case ,

or

, in which case .

The correct choice is .

First, rewrite the quadratic equation in standard form by moving all nonzero terms to the left:

Now factor the quadratic expression into two binomial factors , replacing the question marks with two integers whose product is and whose sum is . These numbers are , so:

or

The solution set is .

First, rewrite the quadratic equation in standard form by FOILing out the product on the left, then collecting all of the terms on the left side:

Use the method to factor the quadratic expression ; we are looking to split the linear term by finding two integers whose sum is and whose product is . These integers are , so:

Set each expression equal to 0 and solve:

or

The solution set is .

can be rewritten as a compound equation:

Solve them separately:

or

The solution set is .

Rewrite the fractions in decimal form:

Now rewrite this as a compound statement and solve each part:

or

The solution set is .

First, rewrite the quadratic equation in standard form by distributing the through the product on the left, then collecting all of the terms on the left side:

Use the method to factor the quadratic expression ; we are looking to split the linear term by finding two integers whose sum is 7 and whose product is . These integers are , so:

Set each expression equal to 0 and solve:

or

The solution set is .

First, rewrite the quadratic equation in standard form by distributing the through the product on the left and collecting all of the terms on the left side:

Use the method to factor the quadratic expression ; we are looking to split the linear term by finding two integers whose sum is and whose product is . These integers are , so:

Set each expression equal to 0 and solve:

or

The solution set is .

First, rewrite the quadratic equation in standard form by FOILing out the product on the left, then collecting all of the terms on the left side:

Use the method to split the middle term into two terms; we want the coefficients to have a sum of 1 and a product of . These numbers are , so we do the following:

Set each expression equal to 0 and solve:

or

The solution set is .

has as its domain the set of values of for which its radicand is nonnegative; that is,

or

Similarly, has as its domain the set of values of for which its radicand is nonnegative; that is,

or

The domain of the sum of two functions is the intersection of the domains of the two individual functions. This intersection is

The domain of is the single value , which is not among the choices.

Examine this diagram:

As indicated by the blue circle in the diagram, the only point on the graph of the function with domain element 4 is , so , and .

As indicated by the red circles in the diagram, there are three points on the graph of the function with range element 4 - namely, , , and . Therefore, . It is unclear whether or without further information.

First, rewrite the quadratic equation in standard form by distributing the through the product on the left, then collecting all of the terms on the left side:

Use the method to factor the quadratic expression ; we are looking to split the linear term by finding two integers whose sum is 7 and whose product is . These integers are , so:

Set each expression equal to 0 and solve:

or

The solution set is .

First, rewrite the quadratic equation in standard form by moving all nonzero terms to the left:

Now factor the quadratic expression into two binomial factors , replacing the question marks with two integers whose product is and whose sum is . These numbers are , so:

or

The solution set is .

First, rewrite the quadratic equation in standard form by FOILing out the product on the left, then collecting all of the terms on the left side:

Use the method to factor the quadratic expression ; we are looking to split the linear term by finding two integers whose sum is and whose product is . These integers are , so:

Set each expression equal to 0 and solve:

or

The solution set is .

can be rewritten as a compound equation:

Solve them separately:

or

The solution set is .