# value of x coordinate

• Sep 23rd 2009, 06:26 AM
mark
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 $\displaystyle 2x + y = 7$ and the curve $\displaystyle y = 4x^2 - 8x - 5$

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

could someone show me the steps to arrive at this answer please?
thanks
• Sep 23rd 2009, 07:10 AM
Quote:

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

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

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

$\displaystyle y=7-2x$ ---1

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

when they intersect , 1=2 .

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

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

Use the quadratic formula to solve for x .
• Sep 23rd 2009, 07:18 AM
mark
yeah thats the problem, i did use the quadratic formula but came up with $\displaystyle 3 \pm \sqrt57$ but i don't understand where the $\displaystyle \frac{1}{4}$ came from?
• Sep 23rd 2009, 07:40 AM
Quote:

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

Oh ok ..

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

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

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

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

so now what's common ?

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