Algebra - How to use the quadratic function

Solve for x.

$-15 + 3x^{2}+4x = -5 - x$

$\frac{-5\pm \sqrt{145}}{6}$

$\frac{-6\pm \sqrt{25}}{5}$

$\frac{-5\pm \sqrt{120}}{3}$

$\frac{5\pm \sqrt{25}}{6}$

Explanation

The quadratic formula is as follows:

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

We will start by finding the values of the coefficients of the given equation, but first we must simplify.

$-15 + 3x^{2}+4x = -5 - x$

Move all the terms to one side and set the equation equal to .

$-15+5 + 3x^{2}+4x+x = 0$

$-10+3x^{2}+5x=0$

Rearrange.

$3x^{2}+5x-10=0$

We can then find the values of the coefficients of the equation:

$3x^{2}+5x-10$

Quadratic equations may be written in the following format:

$ax^{2}+bx+c$

In our case, the values of the coefficients are:

Substitute the coefficient values into the quadratic equation:

$x=\frac{-5\pm \sqrt{5^{2}-4\left ( 3 \right )\left ( -10 \right )}}{2\left ( 3 \right )}$

$x=\frac{-5\pm \sqrt{25-(-120)}}{6}$

After simplifying we are left with:

$x=\frac{-5\pm \sqrt{145}}{6}$

Give the minimum value of the function $f(x) = 2 x^{2} - 5x + 9$ .

$5\frac{7}{8}$

$18\frac{3}{8}$

This function does not have a minimum.

Explanation

This is a quadratic function. The -coordinate of the vertex of the parabola can be determined using the formula $x = - \frac{b}{2a}$ , setting :

$x = - \frac{b}{2a} = - \frac{-5}{2 \cdot 2} =\frac{5}{4}$

Now evaluate the function at $x = \frac{5}{4}$ :

$f(x) = 2 x^{2} - 5x + 9$

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

$= 2 \left (\frac{25}{16} \right ) - 5 \left ( \frac{5}{4} \right )+ 9$

$= \frac{25}{8} - \frac{25}{4} + 9$

$= \frac{25}{8} - \frac{50}{8} + \frac{72}{8} =\frac{47}{8} =5\frac{7}{8}$

Which of the following is the correct solution when is solved using the quadratic equation?

$\frac{10\pm 12}{2}$

$10\pm 12$

$\frac{-10\pm 12}{2}$

$\frac{10\pm \sqrt{56}}{2}$

Explanation

$x=\frac{-(-10)\pm \sqrt{(-10)^2-4(1)(-11)}}{2(1)}$

$=\frac{10\pm \sqrt{100+44}}{2}$

$=\frac{10\pm \sqrt{144}}{2}$

$=\frac{10\pm 12}{2}$

Solve the equation:

$\small 13 = x^2-12x$

$\small \left {-1, 13 \right }$

$\small \left {0\right}$

$\small \left {-13, 1\right }$

$\small \left { 1, 2 \right }$

$\small \left { 2, 4\right }$

Explanation

To solve the quadratic equation, $\small 13= x^{2} -12x$ , we set the equation equal to zero and then factor the quadratic, $\small (x-13)(x+1)=0$ . Because these expressions multiply to equal 0, then it must be that at least one of the expressions equals 0. So we set up the corresponding equations and to obtain the answers and .

Solve for :

$2x^{2}-10x-28=0$

Explanation

To find , we want to factor the quadratic function:

$2x^{2}-10x-28=0$

$2(x^{2}-5x-14)=0$

Solve for :

$x^{2}+12x+35=0$

The solution is undefined.

Explanation

To factor this equation, first find two numbers that multiply to 35 and sum to 12. These numbers are 5 and 7. Split up 12x using these two coefficients: