How do you solve 24+x^2=10x?

Mar 25, 2018

You have to pass $10 x$ to the left hand and equal the quadratic equation to 0
24 + ${x}^{2}$$- 10 x$=0
then you rearrenge it
${x}^{2}$$- 10 x$+24=0

Then you have to think about two numbers that when you times them you get as the answer 24
and when you add them -10

The numbers are -6 and -4
(-6)x(-4)=24
(-6) +(-4)=-10
The final working is :
${x}^{2}$$- 10 x$+24=$\left(x - 6\right) \left(x - 4\right)$

$x - 6 = 0$
$x = 6$

$x - 4 = 0$
$x = 4$

Mar 25, 2018

$x = 6$ or $x = 4$

Explanation:

$24 + {x}^{2} = 10 x$

Put into standard form, $\textcolor{v i o \le t}{a {x}^{2} + b x + c = 0}$

${x}^{2} - 10 x + 24 = 0$

$\downarrow$Factor using criss-cross method of factoring

$1 \textcolor{w h i t e}{X X}$-6

$1 \textcolor{w h i t e}{X X}$-4

$- 4 - 6$

$= - 10$ $\Leftarrow$ same number as our b-value in our rearranged equation.

$\therefore$ $24 + {x}^{2} = 10 x$ is $\textcolor{\mathmr{and} a n \ge}{\text{(x-6)(x-4)}}$

Further on, finding the x-intercepts of $\left(x - 6\right) \left(x - 4\right) = 0$

$x - 6 = 0$ $\textcolor{w h i t e}{X X X X X X}$ and $\textcolor{w h i t e}{X X X X X X}$$x - 4 = 0$

$x = 6$$\textcolor{w h i t e}{X X X X X X X X X X X X X X X X X}$$x = 4$

$\therefore$ the zeros are $\textcolor{b l u e}{6}$ and $\textcolor{b l u e}{4}$.

Mar 25, 2018

$x = 6 \mathmr{and} x = 4$

Explanation:

Here,

$24 + {x}^{2} = 10 x$

$\implies {x}^{2} - 10 x + 24 = 0$

Now,

$\left(- 6\right) \left(- 4\right) = 24 \mathmr{and} \left(- 6\right) + \left(- 4\right) = - 10$

So,

${x}^{2} - 6 x - 4 x + 24 = 0$

$\implies x \left(x - 6\right) - 4 \left(x - 6\right) = 0$

$\implies \left(x - 6\right) \left(x - 4\right) = 0$

$\implies x - 6 = 0 \mathmr{and} x - 4 = 0$

$\implies x = 6 \mathmr{and} x = 4$