Solve for x (2x^2+5x+2)^(1/2)=3

${(2 x^{2} + 5 x + 2)}^{\frac{1}{2}} = 3$

Raise each side of the equation to the power of $2$ to eliminate the fractional exponent on the left side.

${({(2 x^{2} + 5 x + 2)}^{\frac{1}{2}})}^{2} = 3^{2}$

Simplify the exponent.

Simplify the left side.

Simplify ${({(2 x^{2} + 5 x + 2)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(2 x^{2} + 5 x + 2)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(2 x^{2} + 5 x + 2)}^{\frac{1}{2} \cdot 2} = 3^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(2 x^{2} + 5 x + 2)}^{\frac{1}{2} \cdot 2} = 3^{2}$

Rewrite the expression.

${(2 x^{2} + 5 x + 2)}^{1} = 3^{2}$

Simplify.

$2 x^{2} + 5 x + 2 = 3^{2}$

Simplify the right side.

Raise $3$ to the power of $2$ .

$2 x^{2} + 5 x + 2 = 9$

Solve for $x$ .

Subtract $9$ from both sides of the equation.

$2 x^{2} + 5 x + 2 - 9 = 0$

Subtract $9$ from $2$ .

$2 x^{2} + 5 x - 7 = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 7 = - 14$ and whose sum is $b = 5$ .

Factor $5$ out of $5 x$ .

$2 x^{2} + 5 (x) - 7 = 0$

Rewrite $5$ as $- 2$ plus $7$

$2 x^{2} + (- 2 + 7) x - 7 = 0$

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x^{2} - 2 x) + 7 x - 7 = 0$

Factor out the greatest common factor (GCF) from each group.

$2 x (x - 1) + 7 (x - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $x - 1$ .

$(x - 1) (2 x + 7) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 1 = 0$

$2 x + 7 = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $2 x + 7$ equal to $0$ and solve for $x$ .

Set $2 x + 7$ equal to $0$ .

$2 x + 7 = 0$

Solve $2 x + 7 = 0$ for $x$ .

Subtract $7$ from both sides of the equation.

$2 x = - 7$

Divide each term in $2 x = - 7$ by $2$ and simplify.

Divide each term in $2 x = - 7$ by $2$ .

$\frac{2 x}{2} = \frac{- 7}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 7}{2}$

Divide $x$ by $1$ .

$x = \frac{- 7}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{7}{2}$

The final solution is all the values that make $(x - 1) (2 x + 7) = 0$ true.

$x = 1, - \frac{7}{2}$

The result can be shown in multiple forms.

Exact Form:

$x = 1, - \frac{7}{2}$

Decimal Form:

$x = 1, - 3.5$

Mixed Number Form:

$x = 1, - 3 \frac{1}{2}$

Solve for x (2x^2+5x+2)^(1/2)=3

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