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andriy [413]
3 years ago
11

Write the equation in standard form. Then factor the left side of the equation. 3x2 + 19x = 14

Mathematics
2 answers:
solong [7]3 years ago
7 0
3x^2+19x-14=0 Standard form
(3x-2)(x+7)=0 Factored
Airida [17]3 years ago
6 0

Answer:

Standard form of the equation, 3x^2+19x-14=0

Factor of the equation is, (x+7)(3x-2)=0

Step-by-step explanation:

The standard form of the quadratic equation is given as

Ax^2+Bx+C = 0

Given the equation:

3x^2+19x = 14

Subtract 14 from both sides we have;

3x^2+19x-14=0

⇒The standard form of the equation is,  3x^2+19x-14=0

Now, find the factor of the given equation:

3x^2+19x-14=0

⇒3x^2+21x-2x-14=0

⇒3x(x+7)-2(x+7) = 0

⇒(x+7)(3x-2)=0

Therefore, the factor of the equation is, (x+7)(3x-2)=0

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<h2>Hello!</h2>

The answers are:

The possible values for x in the equation, are:

First option, 5\sqrt[3]{3}

Second option,  \sqrt[3]{375}

<h2>Why?</h2>

To solve the problem, we need to remember the following properties of the exponents and roots:

a\sqrt[n]{b}=\sqrt[n]{a^{n}*b} \\\\\sqrt[n]{a^{m} }=a^{\frac{m}{n}}\\\\(a^{b})^{c}=a^{b*c}

Then, we are given the expression:

x^{3}=375

So, finding "x", we have:

x^{3}=375\\\\(x^{3})^{\frac{1}{3} } =(375)^{\frac{1}{3}}\\\\x=\sqrt[3]{375}=\sqrt[3]{125*3}=\sqrt[3]{125}*\sqrt[3]{3}=5\sqrt[3]{3}

Hence, the possible values for x in the equation, are:

First option, 5\sqrt[3]{3}

Second option,  \sqrt[3]{375}

Have a nice day!

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