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All New SAT Math - Calculator Resources

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Flashcards: Systems of Inequalities

Solve the following inequality for $\uptext{x}$ . Round your answer to the nearest tenth.

$\sqrt{x + 7} > x + 1$

2.0 < x < -3.0

2.0 < x

-3.0 < x < 2.0

x = 2.0

$-3.0 \leq x$

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Flashcards: Systems of Inequalities

Solve the following inequality for $\uptext{x}$ . Round your answer to the nearest tenth.

$\sqrt{x + 7} > x + 1$

Your answer:

2.0 < x < -3.0

Your answer:

2.0 < x

Your answer:

x = 2.0

Your answer:

$-3.0 \leq x$

Correct answer:

-3.0 < x < 2.0

Explanation:

The first step is to square each side of the inequality.

$\left( \sqrt{x + 7} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 7 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 7 -\left( x + 7 \right)> x^{2} + 2 x + 1 -\left( x + 7 \right)$

$0 > x^{2} + x - 6$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case a = 1 , b = 1 , and c = -6 .

Now plug these values into the quadratic equation, and we get.

x = 2.0

x = -3.0

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

-3.0 < x < 2.0

Flashcards: Systems of Inequalities

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 10} > x + 1$

$-3.5 \leq x$

2.5 < x

x = 2.5

2.5 < x < -3.5

-3.5 < x < 2.5

Report an Error Copyright Notice

Flashcards: Systems of Inequalities

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 10} > x + 1$

Your answer:

$-3.5 \leq x$

Your answer:

2.5 < x

Your answer:

x = 2.5

Your answer:

2.5 < x < -3.5

Correct answer:

-3.5 < x < 2.5

Explanation:

The first step is to square each side of the inequality.

$\left( \sqrt{x + 10} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 10 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 10 -\left( x + 10 \right)> x^{2} + 2 x + 1 -\left( x + 10 \right)$

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

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case a = 1 , b = 1 , and c = -9 .

Now plug these values into the quadratic equation, and we get.

x = 2.5

x = -3.5

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

-3.5 < x < 2.5

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Flashcards: Systems of Inequalities

Flashcards: Systems of Inequalities

Flashcards: Systems of Inequalities

Flashcards: Systems of Inequalities

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All New SAT Math - Calculator Resources

Flashcards: Systems of Inequalities

Flashcards: Systems of Inequalities

Flashcards: Systems of Inequalities

Flashcards: Systems of Inequalities

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

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