Solve for x square root of 25x^2-40x+16=3

$\sqrt{25 x^{2} - 40 x + 16} = 3$

Step 1

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{25 x^{2} - 40 x + 16}}^{2} = 3^{2}$

Step 2

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{25 x^{2} - 40 x + 16}$ as ${(25 x^{2} - 40 x + 16)}^{\frac{1}{2}}$ .

${({(25 x^{2} - 40 x + 16)}^{\frac{1}{2}})}^{2} = 3^{2}$

Simplify the left side.

Simplify ${({(25 x^{2} - 40 x + 16)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(25 x^{2} - 40 x + 16)}^{\frac{1}{2}})}^{2}$ .

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

${(25 x^{2} - 40 x + 16)}^{\frac{1}{2} \cdot 2} = 3^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(25 x^{2} - 40 x + 16)}^{\frac{1}{2} \cdot 2} = 3^{2}$

Rewrite the expression.

${(25 x^{2} - 40 x + 16)}^{1} = 3^{2}$

Simplify.

$25 x^{2} - 40 x + 16 = 3^{2}$

Simplify the right side.

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

$25 x^{2} - 40 x + 16 = 9$

Step 3

Solve for $x$ .

Subtract $9$ from both sides of the equation.

$25 x^{2} - 40 x + 16 - 9 = 0$

Subtract $9$ from $16$ .

$25 x^{2} - 40 x + 7 = 0$

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 $a \cdot c = 25 \cdot 7 = 175$ and whose sum is $b = - 40$ .

Factor $- 40$ out of $- 40 x$ .

$25 x^{2} - 40 x + 7 = 0$

Rewrite $- 40$ as $- 5$ plus $- 35$

$25 x^{2} + (- 5 - 35) x + 7 = 0$

Apply the distributive property.

$25 x^{2} - 5 x - 35 x + 7 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(25 x^{2} - 5 x) - 35 x + 7 = 0$

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

$5 x (5 x - 1) - 7 (5 x - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $5 x - 1$ .

$(5 x - 1) (5 x - 7) = 0$

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

$5 x - 1 = 0$

$5 x - 7 = 0$

Set $5 x - 1$ equal to $0$ and solve for $x$ .

Set $5 x - 1$ equal to $0$ .

$5 x - 1 = 0$

Solve $5 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$5 x = 1$

Divide each term in $5 x = 1$ by $5$ and simplify.

Divide each term in $5 x = 1$ by $5$ .

$\frac{5 x}{5} = \frac{1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{1}{5}$

Divide $x$ by $1$ .

$x = \frac{1}{5}$

Set $5 x - 7$ equal to $0$ and solve for $x$ .

Set $5 x - 7$ equal to $0$ .

$5 x - 7 = 0$

Solve $5 x - 7 = 0$ for $x$ .

Add $7$ to both sides of the equation.

$5 x = 7$

Divide each term in $5 x = 7$ by $5$ and simplify.

Divide each term in $5 x = 7$ by $5$ .

$\frac{5 x}{5} = \frac{7}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{7}{5}$

Divide $x$ by $1$ .

$x = \frac{7}{5}$

The final solution is all the values that make $(5 x - 1) (5 x - 7) = 0$ true.

$x = \frac{1}{5}, \frac{7}{5}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{5}, \frac{7}{5}$

Decimal Form:

$x = 0.2, 1.4$

Mixed Number Form:

$x = \frac{1}{5}, 1 \frac{2}{5}$

Solve for x square root of 25x^2-40x+16=3

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