Solve 4/(t - 3) + t/2 = - 2 | 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

$$\frac{4}{t-3}+\frac{t}{2}=-2$$

Solve for t

$t = \frac{\sqrt{17} - 1}{2} \approx 1.561552813$
$t=\frac{-\sqrt{17}-1}{2}\approx -2.561552813$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $t$ cannot be equal to $3$ since division by zero is not defined. Multiply both sides of the equation by $2\left(t-3\right)$, the least common multiple of $t-3,2$.

$$2\times 4+\left(t-3\right)t=-4\left(t-3\right)$$

Multiply $2$ and $4$ to get $8$.

$$8+\left(t-3\right)t=-4\left(t-3\right)$$

Use the distributive property to multiply $t-3$ by $t$.

$$8+t^{2}-3t=-4\left(t-3\right)$$

Use the distributive property to multiply $-4$ by $t-3$.

$$8+t^{2}-3t=-4t+12$$

Add $4t$ to both sides.

$$8+t^{2}-3t+4t=12$$

Combine $-3t$ and $4t$ to get $t$.

$$8+t^{2}+t=12$$

Subtract $12$ from both sides.

$$8+t^{2}+t-12=0$$

Subtract $12$ from $8$ to get $-4$.

$$-4+t^{2}+t=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$t^{2}+t-4=0$$

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

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

Square $1$.

$$t=\frac{-1±\sqrt{1-4\left(-4\right)}}{2}$$

Multiply $-4$ times $-4$.

$$t=\frac{-1±\sqrt{1+16}}{2}$$

Add $1$ to $16$.

$$t=\frac{-1±\sqrt{17}}{2}$$

Now solve the equation $t=\frac{-1±\sqrt{17}}{2}$ when $±$ is plus. Add $-1$ to $\sqrt{17}$.

$$t=\frac{\sqrt{17}-1}{2}$$

Now solve the equation $t=\frac{-1±\sqrt{17}}{2}$ when $±$ is minus. Subtract $\sqrt{17}$ from $-1$.

$$t=\frac{-\sqrt{17}-1}{2}$$

The equation is now solved.

$$t=\frac{\sqrt{17}-1}{2}$$ $$t=\frac{-\sqrt{17}-1}{2}$$

Steps for Completing the Square

Variable $t$ cannot be equal to $3$ since division by zero is not defined. Multiply both sides of the equation by $2\left(t-3\right)$, the least common multiple of $t-3,2$.

$$2\times 4+\left(t-3\right)t=-4\left(t-3\right)$$

Multiply $2$ and $4$ to get $8$.

$$8+\left(t-3\right)t=-4\left(t-3\right)$$

Use the distributive property to multiply $t-3$ by $t$.

$$8+t^{2}-3t=-4\left(t-3\right)$$

Use the distributive property to multiply $-4$ by $t-3$.

$$8+t^{2}-3t=-4t+12$$

Add $4t$ to both sides.

$$8+t^{2}-3t+4t=12$$

Combine $-3t$ and $4t$ to get $t$.

$$8+t^{2}+t=12$$

Subtract $8$ from both sides.

$$t^{2}+t=12-8$$

Subtract $8$ from $12$ to get $4$.

$$t^{2}+t=4$$

Divide $1$, the coefficient of the $x$ term, by $2$ to get $\frac{1}{2}$. Then add the square of $\frac{1}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square $\frac{1}{2}$ by squaring both the numerator and the denominator of the fraction.

$$t^{2}+t+\frac{1}{4}=4+\frac{1}{4}$$

Add $4$ to $\frac{1}{4}$.

$$t^{2}+t+\frac{1}{4}=\frac{17}{4}$$

Factor $t^{2}+t+\frac{1}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(t+\frac{1}{2}\right)^{2}=\frac{17}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{1}{2}\right)^{2}}=\sqrt{\frac{17}{4}}$$

Simplify.

$$t+\frac{1}{2}=\frac{\sqrt{17}}{2}$$ $$t+\frac{1}{2}=-\frac{\sqrt{17}}{2}$$

Subtract $\frac{1}{2}$ from both sides of the equation.

$$t=\frac{\sqrt{17}-1}{2}$$ $$t=\frac{-\sqrt{17}-1}{2}$$