Question

algebra 2a semester online practice
complete this assessment to review what youve learned. it will not count
solve the cube root equation (sqrt3{x^3 + 14x^2 + 49x + 20} = x + 5). (1 point)
( circ ) ( x = -35 ) and ( x = -3 )
( circ ) ( x = 3 ) and ( x = 35 )
( circ ) ( x = 5 ) and ( x = 21 )
( circ ) ( x = -21 ) and ( x = -5 )

Explanation:

Step1: Cube both sides

To eliminate the cube root, we cube both sides of the equation \(\sqrt[3]{x^{3}+14x^{2}+49x + 20}=x + 5\). Cubing the left side gives us the expression inside the cube root, and cubing the right side \((x + 5)\) using the formula \((a+b)^3=a^3+3a^2b + 3ab^2+b^3\) (here \(a=x\) and \(b = 5\)) we get \(x^{3}+15x^{2}+75x+125\). So the equation becomes:

$$x^{3}+14x^{2}+49x + 20=x^{3}+15x^{2}+75x+125$$

Step2: Simplify the equation

Subtract \(x^{3}\) from both sides of the equation:

$$14x^{2}+49x + 20=15x^{2}+75x+125$$

Then, move all terms to one side to form a quadratic equation. Subtract \(14x^{2}+49x + 20\) from both sides:

$$0=15x^{2}-14x^{2}+75x - 49x+125 - 20$$

Simplify the like terms:

$$x^{2}+26x + 105 = 0$$

Step3: Factor the quadratic equation

We need to factor \(x^{2}+26x + 105\). We look for two numbers that multiply to \(105\) and add up to \(26\). The numbers are \(21\) and \(5\) since \(21\times5 = 105\) and \(21 + 5=26\). So the factored form is:

$$(x + 21)(x+ 5)=0$$

Step4: Solve for \(x\)

Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\). So we set each factor equal to zero:

• For \(x+21=0\), we get \(x=-21\)
• For \(x + 5=0\), we get \(x=-5\)

We should also check these solutions in the original equation.

Check \(x=-21\):

Left side: \(\sqrt[3]{(-21)^{3}+14\times(-21)^{2}+49\times(-21)+20}=\sqrt[3]{-9261+14\times441-1029 + 20}=\sqrt[3]{-9261 + 6174-1029+20}=\sqrt[3]{-9261+6194}=\sqrt[3]{-3067}\)? Wait, no, let's recalculate:

\((-21)^3=-9261\), \(14\times(-21)^2=14\times441 = 6174\), \(49\times(-21)=-1029\)

So \((-21)^3+14\times(-21)^2+49\times(-21)+20=-9261 + 6174-1029 + 20=-9261+6174=-3087;-3087-1029=-4116;-4116 + 20=-4096\)? Wait, I made a mistake in the previous calculation. Let's do it again:

\((-21)^3=-9261\)

\(14\times(-21)^2=14\times441 = 6174\)

\(49\times(-21)=-1029\)

So sum: \(-9261+6174=-3087\); \(-3087-1029=-4116\); \(-4116 + 20=-4096\)

\(\sqrt[3]{-4096}=-16\)

Right side: \(x + 5=-21 + 5=-16\). So \(x=-21\) is a valid solution.

Check \(x=-5\):

Left side: \(\sqrt[3]{(-5)^{3}+14\times(-5)^{2}+49\times(-5)+20}=\sqrt[3]{-125+14\times25-245 + 20}=\sqrt[3]{-125 + 350-245+20}=\sqrt[3]{(-125-245)+(350 + 20)}=\sqrt[3]{-370+370}=\sqrt[3]{0}=0\)

Right side: \(x + 5=-5 + 5=0\). So \(x=-5\) is also a valid solution.

Answer:

\(x=-21\) and \(x = - 5\) (the option: \(x=-21\) and \(x=-5\))