How to find a solution set - Algebra

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Question

Give all real solutions of the following equation:

$x^{4} +5x^{2} - 36 = 0$

Answer

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} +5x^{2} - 36 = 0$

$u^{2} + 5u - 36 = 0$

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product and sum 5; these integers are .

Substitute back:

$(x^{2}+9)(x^{2}-4) = 0$

The first factor cannot be factored further. The second factor, however, can itself be factored as the difference of squares:

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

Set each factor to zero and solve:

$x^{2}+9= 0$

$x^{2}= -9$

Since no real number squared is equal to a negative number, no real solution presents itself here.

$x-2= 0 \Rightarrow x = 2$

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

The solution set is $\left \{ -2, 2\right \}$ .

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Question

Give all real solutions of the following equation:

$x^{4} +5x^{2} - 36 = 0$

Answer

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} +5x^{2} - 36 = 0$

$u^{2} + 5u - 36 = 0$

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product and sum 5; these integers are .

Substitute back:

$(x^{2}+9)(x^{2}-4) = 0$

The first factor cannot be factored further. The second factor, however, can itself be factored as the difference of squares:

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

Set each factor to zero and solve:

$x^{2}+9= 0$

$x^{2}= -9$

Since no real number squared is equal to a negative number, no real solution presents itself here.

$x-2= 0 \Rightarrow x = 2$

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

The solution set is $\left \{ -2, 2\right \}$ .

Compare your answer with the correct one above

Question

Give all real solutions of the following equation:

$x^{4} +5x^{2} - 36 = 0$

Answer

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} +5x^{2} - 36 = 0$

$u^{2} + 5u - 36 = 0$

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product and sum 5; these integers are .

Substitute back:

$(x^{2}+9)(x^{2}-4) = 0$

The first factor cannot be factored further. The second factor, however, can itself be factored as the difference of squares:

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

Set each factor to zero and solve:

$x^{2}+9= 0$

$x^{2}= -9$

Since no real number squared is equal to a negative number, no real solution presents itself here.

$x-2= 0 \Rightarrow x = 2$

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

The solution set is $\left \{ -2, 2\right \}$ .

Compare your answer with the correct one above

Question

Give all real solutions of the following equation:

$x^{4} +5x^{2} - 36 = 0$

Answer

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} +5x^{2} - 36 = 0$

$u^{2} + 5u - 36 = 0$

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product and sum 5; these integers are .

Substitute back:

$(x^{2}+9)(x^{2}-4) = 0$

The first factor cannot be factored further. The second factor, however, can itself be factored as the difference of squares:

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

Set each factor to zero and solve:

$x^{2}+9= 0$

$x^{2}= -9$

Since no real number squared is equal to a negative number, no real solution presents itself here.

$x-2= 0 \Rightarrow x = 2$

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

The solution set is $\left \{ -2, 2\right \}$ .

Compare your answer with the correct one above

Question

Give all real solutions of the following equation:

$x^{4} +5x^{2} - 36 = 0$

Answer

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} +5x^{2} - 36 = 0$

$u^{2} + 5u - 36 = 0$

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product and sum 5; these integers are .

Substitute back:

$(x^{2}+9)(x^{2}-4) = 0$

The first factor cannot be factored further. The second factor, however, can itself be factored as the difference of squares:

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

Set each factor to zero and solve:

$x^{2}+9= 0$

$x^{2}= -9$

Since no real number squared is equal to a negative number, no real solution presents itself here.

$x-2= 0 \Rightarrow x = 2$

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

The solution set is $\left \{ -2, 2\right \}$ .

Compare your answer with the correct one above

Question

Give all real solutions of the following equation:

$x^{4} +5x^{2} - 36 = 0$

Answer

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} +5x^{2} - 36 = 0$

$u^{2} + 5u - 36 = 0$

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product and sum 5; these integers are .

Substitute back:

$(x^{2}+9)(x^{2}-4) = 0$

The first factor cannot be factored further. The second factor, however, can itself be factored as the difference of squares:

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

Set each factor to zero and solve:

$x^{2}+9= 0$

$x^{2}= -9$

Since no real number squared is equal to a negative number, no real solution presents itself here.

$x-2= 0 \Rightarrow x = 2$

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

The solution set is $\left \{ -2, 2\right \}$ .

Compare your answer with the correct one above

Question

Give all real solutions of the following equation:

$x^{4} - 13x^{2} + 36 = 0$

Answer

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} - 13x^{2} + 36 = 0$

$u^{2} - 13u + 36 = 0$

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product 36 and sum ; these integers are .

Substitute back:

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

These factors can themselves be factored as the difference of squares:

Set each factor to zero and solve:

$x-3 = 0 \Rightarrow x = 3$

$x-2= 0 \Rightarrow x = 2$

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

$x+3 = 0 \Rightarrow x = - 3$

The solution set is $\left \{ -3, -2, 2, 3\right \}$ .

Compare your answer with the correct one above

Question

Give all real solutions of the following equation:

$x^{4} - 13x^{2} + 36 = 0$

Answer

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} - 13x^{2} + 36 = 0$

$u^{2} - 13u + 36 = 0$

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product 36 and sum ; these integers are .

Substitute back:

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

These factors can themselves be factored as the difference of squares:

Set each factor to zero and solve:

$x-3 = 0 \Rightarrow x = 3$

$x-2= 0 \Rightarrow x = 2$

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

$x+3 = 0 \Rightarrow x = - 3$

The solution set is $\left \{ -3, -2, 2, 3\right \}$ .

Compare your answer with the correct one above

Question

Give all real solutions of the following equation:

$x^{4} - 13x^{2} + 36 = 0$

Answer

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} - 13x^{2} + 36 = 0$

$u^{2} - 13u + 36 = 0$

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product 36 and sum ; these integers are .

Substitute back:

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

These factors can themselves be factored as the difference of squares:

Set each factor to zero and solve:

$x-3 = 0 \Rightarrow x = 3$

$x-2= 0 \Rightarrow x = 2$

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

$x+3 = 0 \Rightarrow x = - 3$

The solution set is $\left \{ -3, -2, 2, 3\right \}$ .

Compare your answer with the correct one above

Question

Give all real solutions of the following equation:

$x^{4} - 13x^{2} + 36 = 0$

Answer

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} - 13x^{2} + 36 = 0$

$u^{2} - 13u + 36 = 0$

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product 36 and sum ; these integers are .

Substitute back:

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

These factors can themselves be factored as the difference of squares:

Set each factor to zero and solve:

$x-3 = 0 \Rightarrow x = 3$

$x-2= 0 \Rightarrow x = 2$

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

$x+3 = 0 \Rightarrow x = - 3$

The solution set is $\left \{ -3, -2, 2, 3\right \}$ .

Compare your answer with the correct one above

Question

Give all real solutions of the following equation:

$x^{4} - 13x^{2} + 36 = 0$

Answer

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} - 13x^{2} + 36 = 0$

$u^{2} - 13u + 36 = 0$

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product 36 and sum ; these integers are .

Substitute back:

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

These factors can themselves be factored as the difference of squares:

Set each factor to zero and solve:

$x-3 = 0 \Rightarrow x = 3$

$x-2= 0 \Rightarrow x = 2$

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

$x+3 = 0 \Rightarrow x = - 3$

The solution set is $\left \{ -3, -2, 2, 3\right \}$ .

Compare your answer with the correct one above

Question

Give all real solutions of the following equation:

$x^{4} +5x^{2} - 36 = 0$

Answer

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} +5x^{2} - 36 = 0$

$u^{2} + 5u - 36 = 0$

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product and sum 5; these integers are .

Substitute back:

$(x^{2}+9)(x^{2}-4) = 0$

The first factor cannot be factored further. The second factor, however, can itself be factored as the difference of squares:

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

Set each factor to zero and solve:

$x^{2}+9= 0$

$x^{2}= -9$

Since no real number squared is equal to a negative number, no real solution presents itself here.

$x-2= 0 \Rightarrow x = 2$

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

The solution set is $\left \{ -2, 2\right \}$ .

Compare your answer with the correct one above

Question

Give all real solutions of the following equation:

$x^{4} - 13x^{2} + 36 = 0$

Answer

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} - 13x^{2} + 36 = 0$

$u^{2} - 13u + 36 = 0$

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product 36 and sum ; these integers are .

Substitute back:

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

These factors can themselves be factored as the difference of squares:

Set each factor to zero and solve:

$x-3 = 0 \Rightarrow x = 3$

$x-2= 0 \Rightarrow x = 2$

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

$x+3 = 0 \Rightarrow x = - 3$

The solution set is $\left \{ -3, -2, 2, 3\right \}$ .

Compare your answer with the correct one above

Question

Give all real solutions of the following equation:

$x^{4} +5x^{2} - 36 = 0$

Answer

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} +5x^{2} - 36 = 0$

$u^{2} + 5u - 36 = 0$

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product and sum 5; these integers are .

Substitute back:

$(x^{2}+9)(x^{2}-4) = 0$

The first factor cannot be factored further. The second factor, however, can itself be factored as the difference of squares:

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

Set each factor to zero and solve:

$x^{2}+9= 0$

$x^{2}= -9$

Since no real number squared is equal to a negative number, no real solution presents itself here.

$x-2= 0 \Rightarrow x = 2$

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

The solution set is $\left \{ -2, 2\right \}$ .

Compare your answer with the correct one above

Question

Give all real solutions of the following equation:

$x^{4} - 13x^{2} + 36 = 0$

Answer

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} - 13x^{2} + 36 = 0$

$u^{2} - 13u + 36 = 0$

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product 36 and sum ; these integers are .

Substitute back:

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

These factors can themselves be factored as the difference of squares:

Set each factor to zero and solve:

$x-3 = 0 \Rightarrow x = 3$

$x-2= 0 \Rightarrow x = 2$

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

$x+3 = 0 \Rightarrow x = - 3$

The solution set is $\left \{ -3, -2, 2, 3\right \}$ .

Compare your answer with the correct one above

Question

Give all real solutions of the following equation:

$x^{4} - 13x^{2} + 36 = 0$

Answer

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} - 13x^{2} + 36 = 0$

$u^{2} - 13u + 36 = 0$

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product 36 and sum ; these integers are .

Substitute back:

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

These factors can themselves be factored as the difference of squares:

Set each factor to zero and solve:

$x-3 = 0 \Rightarrow x = 3$

$x-2= 0 \Rightarrow x = 2$

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

$x+3 = 0 \Rightarrow x = - 3$

The solution set is $\left \{ -3, -2, 2, 3\right \}$ .

Compare your answer with the correct one above

Question

Give all real solutions of the following equation:

$x^{4} - 13x^{2} + 36 = 0$

Answer

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} - 13x^{2} + 36 = 0$

$u^{2} - 13u + 36 = 0$

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product 36 and sum ; these integers are .

Substitute back:

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

These factors can themselves be factored as the difference of squares:

Set each factor to zero and solve:

$x-3 = 0 \Rightarrow x = 3$

$x-2= 0 \Rightarrow x = 2$

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

$x+3 = 0 \Rightarrow x = - 3$

The solution set is $\left \{ -3, -2, 2, 3\right \}$ .

Compare your answer with the correct one above

Question

Give all real solutions of the following equation:

$x^{4} +5x^{2} - 36 = 0$

Answer

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} +5x^{2} - 36 = 0$

$u^{2} + 5u - 36 = 0$

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product and sum 5; these integers are .

Substitute back:

$(x^{2}+9)(x^{2}-4) = 0$

The first factor cannot be factored further. The second factor, however, can itself be factored as the difference of squares:

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

Set each factor to zero and solve:

$x^{2}+9= 0$

$x^{2}= -9$

Since no real number squared is equal to a negative number, no real solution presents itself here.

$x-2= 0 \Rightarrow x = 2$

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

The solution set is $\left \{ -2, 2\right \}$ .

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Question

Solve for :

$x = \sqrt{2x-1} +8$

Answer

Isolate the radical, square both sides, and solve the resulting quadratic equation.

$x = \sqrt{2x-1} +8$

$x - 8 = \sqrt{2x-1} +8 -8$

$x - 8 = \sqrt{2x-1}$

$\left (x -8 \right )^{2} =\left ( \sqrt{2x-1} \right )^{2}$

$x^{2} - 2 \cdot x \cdot 8 + 8^{2} = 2x - 1$

$x^{2} - 16 x + 64 = 2x - 1$

$x^{2} - 16 x + 64 - 2x + 1 = 2x - 1 - 2x + 1$

$x^{2} - 18 x + 65 = 0$

Factor the expression at left by finding two integers whose product is 65 and whose sum is ; they are .

Set each linear binomial to 0 and solve for to find the possible solutions.

or

Substitute each for .

$x = \sqrt{2x-1} +8$

$5 = \sqrt{2 \cdot 5-1} +8$

$5 = \sqrt{9} +8$

This is a false statement, so 5 is a false "solution".

$x = \sqrt{2x-1} +8$

$13 = \sqrt{2 \cdot 13-1} +8$

$13= \sqrt{25} +8$

This is a true statement, so 13 is the only solution of the equation.

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Question

Solve this system of equations.

Answer

Equation 1:

Equation 2:

Equation 3:

Adding the terms of the first and second equations together will yield .

Then, add that to the third equation so that the y and z terms are eliminated. You will get .

This tells us that x = 1. Plug this x = 1 back into the systems of equations.

Now, we can do the rest of the problem by using the substitution method. We'll take the third equation and use it to solve for y.

Plug this y-equation into the first equation (or second equation; it doesn't matter) to solve for z.

$z=-\frac{5}{3}$

We can use this z value to find y

$y=-3\left ( -\frac{5}{3}\right )-3$

So the solution set is x = 1, y = 2, and z = –5/3.

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