Solving Quadratic Equations - Math

Card 0 of 172

Question

Solve the quadratic equation using any method:

Answer

Use the quadratic formula to solve:

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x = \frac{-(c) \pm \sqrt{(c)^2-4(a)(b)}}{2(b)}$

$x = \frac{-c \pm \sqrt{c^2-4ab}}{2b}$

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Question

$f(x)=x^{2}-4x+7$

Find the vertex of the parabola by completing the square.

Answer

To find the vertex of a parabola, we must put the equation into the vertex form:

$\small f(x)=a[b(x-h)]^{2}+k$

The vertex can then be found with the coordinates (h, k).

To put the parabola's equation into vertex form, you have to complete the square. Completing the square just means adding the same number to both sides of the equation -- which, remember, doesn't change the value of the equation -- in order to create a perfect square.

Start with the original equation:

$f(x)=x^{2}-4x+7$

Put all of the terms on one side:

$f(x)-7=x^{2}-4x$

Now we know that we have to add something to both sides in order to create a perfect square:

$f(x)+7+\square =x^{2}-4x+\square$

In this case, we need to add 4 on both sides so that the right-hand side of the equation factors neatly.

$f(x)-7+4=x^{2}-4x+4$

Now we factor:

$f(x)-3=(x-2)^{2}$

Once we isolate , we have the equation in vertex form:

$f(x)=(x-2)^{2}+3$

Thus, the parabola's vertex can be found at $\small (2,3)$ .

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Question

Complete the square:

Answer

Begin by dividing the equation by and subtracting from each side:

$x^2+\frac{3}{2}x = -1$

Square the value in front of the and add to each side:

$x^2+\frac{3}{2}x +\frac{9}{16} = -1+\frac{9}{16}$

Factor the left side of the equation:

$(x+\frac{3}{4})^2 = \frac{-7}{16}$

Take the square root of both sides and simplify:

$x+\frac{3}{4} = \frac{\pm \sqrt{-7}}{4}$

$x = \frac{-3 \pm i\sqrt{7}}{4}$

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Question

Use factoring to solve the quadratic equation:

Answer

Factor and solve:

Factor like terms:

Combine like terms:

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

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Question

Complete the square:

Answer

Begin by dividing the equation by and adding to each side:

$n^2+\frac{1}{2}n = \frac{21}{2}$

Square the value in front of the and add to each side:

$n^2+\frac{1}{2}n + \frac{1}{16} = \frac{21}{2} + \frac{1}{16}$

Factor the left side of the equation:

$(n+\frac{1}{4})^2 = \frac{169}{16}$

Take the square root of both sides and simplify:

$n+\frac{1}{4} = \frac{\pm 13}{4}$

$n = \frac{-1 \pm 13}{4}$

$n=-\frac{7}{2},3$

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Question

Complete the square:

Answer

Begin by dividing the equation by and subtracting $\frac{2}{3}$ from each side:

$x^2+\frac{4}{3}x = -\frac{2}{3}$

Square the value in front of the and add to each side:

$x^2+\frac{4}{3}x +\frac{4}{9} = -\frac{2}{3}+\frac{4}{9}$

Factor the left side of the equation:

$(x+\frac{2}{3})^2 = \frac{-2}{9}$

Take the square root of both sides and simplify:

$x+\frac{2}{3} = \frac{\pm \sqrt{-2}}{3}$

$x = \frac{-2 \pm i\sqrt{2}}{3}$

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Question

Find the zeros.

$f(t)=t^{2}-4t-21$

Answer

Factor the equation to . Set both equal to zero and you get and . Remember, the zeros of an equation are wherever the function crosses the -axis.

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Question

Find the zeros.

$f(x)=x^{2}-10x$

Answer

Factor out an from the equation so that you have . Set and equal to . Your roots are and .

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Question

Find the zeros.

$f(x)=\frac{1}{5}(x+9)^{2}(x^{2}-4)$

Answer

Set equal to zero and you get . Set equal to zero as well and you get and because when you take a square root, your answer will be positive and negative.

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Question

Find the zeros.

$f(x)=27x^{2}-45x-18$

Answer

Factor out a from the entire equation. After that, you get . Factor the expression to . Set both of those equal to zero and your answers are $-\frac{1}{3}$ and .

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Question

Find the zeros.

$f(x)=25x^{2}-81$

Answer

This expression is the difference of perfect squares. Therefore, it factors to. Set both of those equal to zero and your answers are $\frac{9}{5}$ and $-\frac{9}{5}$ .

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Question

Find the zeros.

Answer

Factor the equation to . Set both equal to and you get and $\frac{1}{4}$ .

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Question

Find the zeros.

$f(x)=\frac{1}{4}x^2 +\frac{5}{4}x+\frac{3}{2}$

Answer

Factor a $\frac{1}{4}$ out of the quation to get

$\frac{1}{4}\left ( x^{2}+5x+6 \right )$

which can be further factored to

$\frac{1}{4}(x+3)(x+2)$ .

Set the last two expressions equal to zero and you get and .

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Question

Find the zeros.

Answer

Set each expression equal to zero and you get 0 and 6.

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Question

Find the zeros.

Answer

Set both expressions equal to . The first factor yields . The second factor gives you .

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Question

Find the zeros.

Answer

Set both expressions to and you get and .

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Question

Solve the following equation by factoring.

Answer

We can factor by determining the terms that will multiply to –8 and add to +7.

$8*-1=-8\ and\ 8+(-1)=7$

Our factors are +8 and –1.

Now we can set each factor equal to zero and solve for the root.

$x+8=0\ or\ x-1=0$

$x=-8\ or\ x=1$

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Question

Solve the following equation by factoring.

Answer

We know that one term has a coefficient of 2 and that our factors must multiply to –10.

$2*-5=-10\ and\ 2+2(-5)=-8$

Our factors are +2 and –5.

Now we can set each factor equal to zero and solve for the root.

$2x+2=0\ or\ x-5=0$

$x=-1\ or\ 5$

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Question

Solve the following equation by factoring.

Answer

First, we can factor an term out of all of the values.

We can factor remaining polynomial by determining the terms that will multiply to +4 and add to +4.

$2*2=4\ and\ 2+2=4$

Our factors are +2 and +2.

Now we can set each factor equal to zero and solve for the root.

$x=0\ or\ (x+2)^2=0$

$x=0\ or\ -2$

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Question

Find the sum of the solutions to:

$\frac{-6}{x^2} + \frac{1}{x} + 1 = 0$

Answer

Multiply both sides of the equation by , to get

This can be factored into the form

So we must solve

and

to get the solutions.

The solutions are:

and their sum is .

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