# Math Help - value of x coordinate

1. ## value of x coordinate

hi, the question i've got is:
find the exact value of the x coordinate at the point of intersection of-

the line $2x + y = 7$ and the curve $y = 4x^2 - 8x - 5$

the answer is $\frac{1}{4} (3 + \sqrt{57})$ and $\frac{1}{4}(3 - \sqrt{57})$

could someone show me the steps to arrive at this answer please?
thanks

2. Originally Posted by mark
hi, the question i've got is:
find the exact value of the x coordinate at the point of intersection of-

the line $2x + y = 7$ and the curve $y = 4x^2 - 8x - 5$

the answer is $\frac{1}{4} (3 + \sqrt{57})$ and $\frac{1}{4}(3 - \sqrt{57})$

could someone show me the steps to arrive at this answer please?
thanks
$y=7-2x$ ---1

$y = 4x^2 - 8x - 5$ --- 2

when they intersect , 1=2 .

$7-2x=4x^2-8x-5$

$4x^2-6x-12=0$

Use the quadratic formula to solve for x .

3. yeah thats the problem, i did use the quadratic formula but came up with $3 \pm \sqrt57$ but i don't understand where the $\frac{1}{4}$ came from?

4. Originally Posted by mark
yeah thats the problem, i did use the quadratic formula but came up with $3 \pm \sqrt57$ but i don't understand where the $\frac{1}{4}$ came from?
Oh ok ..

$x=\frac{-(-6)\pm\sqrt{36-4(4)(-12)}}{2(4)}$

$=\frac{6\pm\sqrt{228}}{8}$

$=\frac{6\pm2\sqrt{57}}{8}$

$=\frac{3}{4}\pm\frac{\sqrt{57}}{4}$

so now what's common ?

$
=\frac{1}{4}(3\pm\sqrt{57})
$