Which expression is equivalent to -6 (3x

BigorU [14]

If u =5 and c= -3 then cu = -3+5 = 2

3 0

3 years ago

Write down the quadratic equation whose roots are $x = -7$ and $x = 1,$ and the coefficient of $x^2$ is 1. Enter your answer in

pav-90 [236]

<h2>Steps:</h2>

So firstly, since we know that the coefficient of x² is 1, this means that this is our base equation:

y = x² + bx + c

Now, since we know that the roots are -7 and 1, set y = 0 and set x = -7 and 1 and simplify:

$0=(-7)^2+b(-7)+c\\0=49-7b+c\\-49=-7b+c\\\\0=1^2+b(1)+c\\0=1+b+c\\-1=b+c\\\\-49=-7b+c\\-1=b+c$

Now with this, we can set up a system of equations to solve for b and c. For this, I will be using the elimination method. For this, subtract the 2 equations:

$\begin{alignedat}{2}-49&=-7b+c\\-(-1&=b+c)\\-48&=-8b\end{alignedat}$

Now that the c variable has been eliminated we can solve for b. For this, divide both sides by -8 and your first part of your answer is b = 6.

Now that we know the value of b, plug it into either equation to solve for c:

$-49=-7(6)+c\\-49=-42+c\\-7=c\\\\-1=6+c\\-7=c$

<h2>Answer:</h2>

Putting it together, your final answer is x² + 6x - 7 = 0.

4 0

3 years ago

Help me plsssssssss its easy and show work please

babymother [125]

Answer:

» For x-intercept, y is zero

 y = x² - 12x + 32 

${ \tt{0 = {x}^{2} - 12x + 32 }} \\ { \tt{ {x}^{2} - 12x + 32 = 0 }} \\ { \tt{(x - 8)(x - 4) = 0}} \\ { \tt{x = 8 \: \: and \: \: 4}} \\ { \mathfrak{x - intercept : \: \: }}{ \boxed{ \tt{ \: (8, \: 0) \: \: and \: \: (4, \: 0) \: \: }}}$

 y = x² - 10x +24 :

${ \tt{0 = {x}^{2} - 10x + 24}} \\ { \tt{ {x}^{2} - 10x + 24 = 0}} \\ { \tt{(x - 6)(x - 4) = 0}} \\ { \tt{x = 6 \: \: and \: \: 4}} \\ { \mathfrak{x - intercept : \: \: }}{ \boxed{ \tt{ \: (6, \: 0) \: \: and \: \: (4, \: 0) \: \: }}}$

 y = x² + 10x + 16 :

${ \tt{0 = {x}^{2} + 10x + 16}} \\ { \tt{ {x}^{2} + 10x + 16 = 0 }} \\ { \tt{(x + 2)(x + 8) = 0}} \\ { \tt{x = - 2 \: \: and \: \: - 8}} \\ { \mathfrak{x - intercept : \: \: }}{ \boxed{ \tt{ \: ( - 2, \: 0) \: \: and \: \: ( - 8, \: 0) \: \: }}}$

 y = x² - 25 :

${ \tt{0 = {x}^{2} - 25}} \\ { \tt{ {x}^{2} = 25}} \\ { \tt{ \sqrt{ {x}^{2} } = \sqrt{25} }} \\ { \tt{x = \pm5}} \\ { \mathfrak{x - intercept : \: \: }}{ \boxed{ \tt{ \: (5, \: 0) \: \: and \: \: ( - 5, \: 0) \: \: }}}$

 y = x² + 8x + 12 :

${ \tt{0 = {x}^{2} + 8x + 12 }} \\ { \tt{ {x}^{2} + 8x + 12 = 0 }} \\{ \tt{(x + 2)(x + 6) = 0}} \\ { \mathfrak{x - intercept : \: \: }}{ \boxed{ \tt{ \: ( - 2, \: 0) \: \: and \: \: ( - 6, \: 0) \: \: }}}$