Solve P(x) = 5x ^ 2 - 3x + 7 A at ac = 1; o1=1* o1 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$P ( x ) = 5 x ^ { 2 } - 3 x + 7 x = 1$$

Solve for x

$x=-1$
$x=\frac{1}{5}=0.2$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Combine $-3x$ and $7x$ to get $4x$.

$$5x^{2}+4x=1$$

Subtract $1$ from both sides.

$$5x^{2}+4x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $5x^{2}+ax+bx-1$. To find $a$ and $b$, set up a system to be solved.

$$a+b=4$$ $$ab=5\left(-1\right)=-5$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1$$ $$b=5$$

Rewrite $5x^{2}+4x-1$ as $\left(5x^{2}-x\right)+\left(5x-1\right)$.

$$\left(5x^{2}-x\right)+\left(5x-1\right)$$

Factor out $x$ in $5x^{2}-x$.

$$x\left(5x-1\right)+5x-1$$

Factor out common term $5x-1$ by using distributive property.

$$\left(5x-1\right)\left(x+1\right)$$

To find equation solutions, solve $5x-1=0$ and $x+1=0$.

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

Steps Using the Quadratic Formula

Combine $-3x$ and $7x$ to get $4x$.

$$5x^{2}+4x=1$$

Subtract $1$ from both sides.

$$5x^{2}+4x-1=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $5$ for $a$, $4$ for $b$, and $-1$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-4±\sqrt{4^{2}-4\times 5\left(-1\right)}}{2\times 5}$$

Square $4$.

$$x=\frac{-4±\sqrt{16-4\times 5\left(-1\right)}}{2\times 5}$$

Multiply $-4$ times $5$.

$$x=\frac{-4±\sqrt{16-20\left(-1\right)}}{2\times 5}$$

Multiply $-20$ times $-1$.

$$x=\frac{-4±\sqrt{16+20}}{2\times 5}$$

Add $16$ to $20$.

$$x=\frac{-4±\sqrt{36}}{2\times 5}$$

Take the square root of $36$.

$$x=\frac{-4±6}{2\times 5}$$

Multiply $2$ times $5$.

$$x=\frac{-4±6}{10}$$

Now solve the equation $x=\frac{-4±6}{10}$ when $±$ is plus. Add $-4$ to $6$.

$$x=\frac{2}{10}$$

Reduce the fraction $\frac{2}{10}$ to lowest terms by extracting and canceling out $2$.

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

Now solve the equation $x=\frac{-4±6}{10}$ when $±$ is minus. Subtract $6$ from $-4$.

$$x=-\frac{10}{10}$$

Divide $-10$ by $10$.

$$x=-1$$

The equation is now solved.

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

Steps for Completing the Square

Combine $-3x$ and $7x$ to get $4x$.

$$5x^{2}+4x=1$$

Divide both sides by $5$.

$$\frac{5x^{2}+4x}{5}=\frac{1}{5}$$

Dividing by $5$ undoes the multiplication by $5$.

$$x^{2}+\frac{4}{5}x=\frac{1}{5}$$

Divide $\frac{4}{5}$, the coefficient of the $x$ term, by $2$ to get $\frac{2}{5}$. Then add the square of $\frac{2}{5}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{4}{5}x+\left(\frac{2}{5}\right)^{2}=\frac{1}{5}+\left(\frac{2}{5}\right)^{2}$$

Square $\frac{2}{5}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=\frac{1}{5}+\frac{4}{25}$$

Add $\frac{1}{5}$ to $\frac{4}{25}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=\frac{9}{25}$$

Factor $x^{2}+\frac{4}{5}x+\frac{4}{25}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{2}{5}\right)^{2}=\frac{9}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{2}{5}\right)^{2}}=\sqrt{\frac{9}{25}}$$

Simplify.

$$x+\frac{2}{5}=\frac{3}{5}$$ $$x+\frac{2}{5}=-\frac{3}{5}$$

Subtract $\frac{2}{5}$ from both sides of the equation.

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