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3 years ago

Suppose x = 7 is a solution to the equation 4x - 2(x + a) = 8. Find the value of a that makes the equation true

Mathematics

2 answers:

Sonja [21]3 years ago

8 0

Answer:

a = 7

Step-by-step explanation:

mr Goodwill [35]3 years ago

4 0

Answer:

$a=3$

Step-by-step explanation:

So we have the equation:

$4x-2(x+a)=8$

And we want to find a such that x = 7 is a possible solution.

To do so, substitute 7 for x. Thus:

$4(7)-2(7+a)=8$

Multiply the left and distribute the right:

$28-14-2a=8$

Subtract:

$14-2a=8$

Subtract 14 from both sides:

$-2a=-6$

Divide by -2: "

$a=3$

So, the value of a is 3.

And our equation would be:

$4x-2(x+3)=8$

And we're done!

Edit: Incorrect Answer

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damaskus [11]

<u>ANSWER: </u>

x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

<u>SOLUTION:</u>

Given, $f(x)=x^{2}+12 x+24$ -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

$X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here, for (1) a = 1, b= 12 and c = 24

$X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}$

$\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}$

Hence the x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

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