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PSAT Math : Equations / Inequalities

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Example Questions

Example Question #2 : Systems Of Inequalities

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

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

Possible Answers:

0.0 < x

x = 0.0

-1.0 < x < 0.0

$-1.0 \leq x$

0.0 < x < -1.0

Correct answer:

-1.0 < x < 0.0

Explanation:

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

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

Now simplify each side.

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

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

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

$0 > x^{2} + x$

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 = 0 .

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

x = 0.0

x = -1.0

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.

-1.0 < x < 0.0

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Example Question #3 : Systems Of Inequalities

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

$\sqrt{x + 6} > x + 1$

Possible Answers:

1.8 < x

x = 1.8

-2.8 < x < 1.8

$-2.8 \leq x$

1.8 < x < -2.8

Correct answer:

-2.8 < x < 1.8

Explanation:

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

$\left( \sqrt{x + 6} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 6 > x^{2} + 2 x + 1$

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

$x + 6 -\left( x + 6 \right)> x^{2} + 2 x + 1 -\left( x + 6 \right)$

$0 > x^{2} + x - 5$

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 = -5 .

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

x = 1.8

x = -2.8

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.

-2.8 < x < 1.8

Report an Error

Example Question #1 : Inequalities And Absolute Value

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

$\sqrt{x + 35} > x + 1$

Possible Answers:

-6.4 < x < 5.4

$-6.4 \leq x$

x = 5.4

5.4 < x

5.4 < x < -6.4

Correct answer:

-6.4 < x < 5.4

Explanation:

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

$\left( \sqrt{x + 35} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 35 > x^{2} + 2 x + 1$

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

$x + 35 -\left( x + 35 \right)> x^{2} + 2 x + 1 -\left( x + 35 \right)$

$0 > x^{2} + x - 34$

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 = -34 .

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

x = 5.4

x = -6.4

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.

-6.4 < x < 5.4

Report an Error

Example Question #2 : Inequalities And Absolute Value

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

$\sqrt{x + 17} > x + 1$

Possible Answers:

3.5 < x

$-4.5 \leq x$

3.5 < x < -4.5

x = 3.5

-4.5 < x < 3.5

Correct answer:

-4.5 < x < 3.5

Explanation:

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

$\left( \sqrt{x + 17} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 17 > x^{2} + 2 x + 1$

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

$x + 17 -\left( x + 17 \right)> x^{2} + 2 x + 1 -\left( x + 17 \right)$

$0 > x^{2} + x - 16$

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 = -16 .

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

x = 3.5

x = -4.5

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.

-4.5 < x < 3.5

Report an Error

Example Question #1 : Solve One Variable Linear Equations And Inequalities: Ccss.Math.Content.Hsa Rei.B.3

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

$\sqrt{x + 8} > x + 1$

Possible Answers:

-3.2 < x < 2.2

x = 2.2

$-3.2 \leq x$

2.2 < x

2.2 < x < -3.2

Correct answer:

-3.2 < x < 2.2

Explanation:

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

$\left( \sqrt{x + 8} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 8 > x^{2} + 2 x + 1$

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

$x + 8 -\left( x + 8 \right)> x^{2} + 2 x + 1 -\left( x + 8 \right)$

$0 > x^{2} + x - 7$

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 = -7 .

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

x = 2.2

x = -3.2

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.

-3.2 < x < 2.2

Report an Error

Example Question #491 : High School: Algebra

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

$\sqrt{x + 15} > x + 1$

Possible Answers:

3.3 < x < -4.3

x = 3.3

3.3 < x

-4.3 < x < 3.3

$-4.3 \leq x$

Correct answer:

-4.3 < x < 3.3

Explanation:

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

$\left( \sqrt{x + 15} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 15 > x^{2} + 2 x + 1$

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

$x + 15 -\left( x + 15 \right)> x^{2} + 2 x + 1 -\left( x + 15 \right)$

$0 > x^{2} + x - 14$

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 = -14 .

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

x = 3.3

x = -4.3

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.

-4.3 < x < 3.3

Report an Error

Example Question #191 : Equations / Inequalities

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

$\sqrt{x + 16} > x + 1$

Possible Answers:

3.4 < x < -4.4

$-4.4 \leq x$

-4.4 < x < 3.4

3.4 < x

x = 3.4

Correct answer:

-4.4 < x < 3.4

Explanation:

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

$\left( \sqrt{x + 16} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 16 > x^{2} + 2 x + 1$

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

$x + 16 -\left( x + 16 \right)> x^{2} + 2 x + 1 -\left( x + 16 \right)$

$0 > x^{2} + x - 15$

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 = -15 .

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

x = 3.4

x = -4.4

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.

-4.4 < x < 3.4

Report an Error

Example Question #192 : Equations / Inequalities

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

$\sqrt{x + 34} > x + 1$

Possible Answers:

x = 5.3

-6.3 < x < 5.3

$-6.3 \leq x$

5.3 < x < -6.3

5.3 < x

Correct answer:

-6.3 < x < 5.3

Explanation:

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

$\left( \sqrt{x + 34} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 34 > x^{2} + 2 x + 1$

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

$x + 34 -\left( x + 34 \right)> x^{2} + 2 x + 1 -\left( x + 34 \right)$

$0 > x^{2} + x - 33$

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 = -33 .

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

x = 5.3

x = -6.3

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.

-6.3 < x < 5.3

Report an Error

Example Question #193 : Equations / Inequalities

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

$\sqrt{x + 19} > x + 1$

Possible Answers:

x = 3.8

3.8 < x < -4.8

-4.8 < x < 3.8

3.8 < x

$-4.8 \leq x$

Correct answer:

-4.8 < x < 3.8

Explanation:

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

$\left( \sqrt{x + 19} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 19 > x^{2} + 2 x + 1$

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

$x + 19 -\left( x + 19 \right)> x^{2} + 2 x + 1 -\left( x + 19 \right)$

$0 > x^{2} + x - 18$

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 = -18 .

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

x = 3.8

x = -4.8

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.

-4.8 < x < 3.8

Report an Error

Example Question #1 : Calculate Maximum Or Minimum Of Quadratic By Completing The Square: Ccss.Math.Content.Hsa Sse.B.3b

Complete the square to calculate the maximum or minimum point of the given function.

x^2+4x+7

Possible Answers:

$\text{Minimum}=(2,3)$

$\text{Minimum}=(-2,-3)$

$\text{Minimum}=(-2,3)$

$\text{Maximum}=(-2,-3)$

$\text{Maximum}=(-2,3)$

Correct answer:

$\text{Minimum}=(-2,3)$

Explanation:

Completing the square method uses the concept of perfect squares. Recall that a perfect square is in the form,

$(ax\pm b)^2=(ax\pm b)(ax\pm b)$

where when multiplied out,

a^2x^2+2abx+b^2

the middle term coefficient, when divided by two and squared, results in the coefficient of the last term.

$2ab=\left(\frac{2ab}{2} \right )^2=b^2$

Complete the square for this particular function is as follows.

x^2+4x+7

First identify the middle term coefficient.

$\text{Middle Term Coefficient}= 4$

Now divide the middle term coefficient by two.

$\frac{4}{2}=2$

From here write the function with the perfect square. Remember when adding the new squared term, add it to both sides to keep the equation balanced.

$x^2+4x+7=0\Rightarrow (x+2)^2+2^2+7=0+2^2$

When simplified the new function is,

$\\(x+2)^2+4+7=4 \\(x+2)^2+11=4 \\(x+2)^2=-7$

Since the x^2 term is positive, the parabola will be opening up. This means that the function has a minimum value at the vertex. To find the value of the vertex set the inside portion of the binomial equal to zero and solve.

$\\x+2=0 \\x+2-2=0-2 \\x=-2$

From here, substitute the the value into the original function.

x^2+4x+7

$\\y=(-2)^2+4(-2)+7 \\y=4-8+7 \\y=3$

Therefore the minimum value occurs at the point (-2,3) .

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PSAT Math : Equations / Inequalities

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #2 : Systems Of Inequalities

Example Question #3 : Systems Of Inequalities

Example Question #1 : Inequalities And Absolute Value

Example Question #2 : Inequalities And Absolute Value

Example Question #1 : Solve One Variable Linear Equations And Inequalities: Ccss.Math.Content.Hsa Rei.B.3

Example Question #491 : High School: Algebra

Example Question #191 : Equations / Inequalities

Example Question #192 : Equations / Inequalities

Example Question #193 : Equations / Inequalities

Example Question #1 : Calculate Maximum Or Minimum Of Quadratic By Completing The Square: Ccss.Math.Content.Hsa Sse.B.3b

All PSAT Math Resources

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