find the coordinates of the points at which the line with equation y= x - 4 intersects curve with equation y^2 = 2x^2 - 17.

Question

simultaneous equation problem .HELP ME

Answer 1

y = x-4, and
y^2 = 2x^2 - 17

(x-4)^2 = 2x^2 - 17

x^2 -8x + 16 = 2x^2 - 17

0 = x^2 + 8x - 33

0 = (x + 11)(x - 3)

x = -11 or x = 3

y = x-4 = -15 or -1

Solutions: (-11,-15) and (3,-1)

Answer 2

Hi Rodela,

y=x-4......equation-1

y^2=2x^2-17.......equation2

substitute the value of y from eqn 1 to eqn 2,

we get, (x-4)^2=2x^2-17

x^2-8x+16=2x^2-17

x^2+8x-33=0

x=-b+Sqroot ofb^2-4ac 

           2a

         or x=-b -sqroot of b^2-4ac

                         2a

 

Substituting the value from quadratic equation, a=1,b=8, c=-33,

x=-8+sqroot of(8^2-4*1*-33) /2a 

= -8++sqroot of (64+132)/2

=-8+sqroot of (296)/2=-8+17.40/2=9.40/2=4.70   

or                                   

x=-8-sqroot of (8^2-4*1*-33) /2*1

= -8-sqroot of (64+132)/2=-8-sqrt of (296)/2=-8-17.40/2=-25.40/2=-12.7

substituting the value of x in eq-1,if x=4.70, y=x-4=4.70-4=0.70

if x=-12.7, y=-12.7-4=-16.7

 (4.7,0.7),(-12.7,-16.7)

check: (4.7,0.7) is a solution or not

LHS of eq 2,y^2=(0.7)^2=0.49

RHS of eq 2=2x^2-17=2(4.7)^2-17=2*22.09-17=44.18-17=27.18

LHS is not = to RHS. so it is not a solution.

same way check for (-12.7,-16.7)

 

Meena from strongsville, OH

 

 

Answer 3

Hi Rodela;

y= x - 4

y^2 = 2x^2 - 17.

Let's take the second equation and replace y with x-4...

(x-4)²=2x²-17

Let's FOIL the (x-4)(x-4)...

FIRST...(x)(x)=x²

OUTER...(-4)(x)=-4x

INNER...(-4)(x)=-4x

LAST...(-4)(-4)=16

x²-4x-4x+16

x²-8x+16

x²-8x+16=2x²-17

Let's subtract x² from both sides...

x²-x²-8x+16=2x²-x²-17

-8x+16=x²-17

Let's add 8x to both sides...

-8x+8x+16=x²+8x-17

16=x²+8x-17

Let's subtract 16 from both sides...

16-16=x²+8x-17-16

0=x²+8x-33

For the FOIL...

FIRST must be (x)(x)=x²

OUTER and INNER must add-up to 8x

LAST must be (3)(11) or (11)(3) and only one number must be negative to render the sum of +8 and product of -33.

(x+11)(x-3)=0

Either or both parenthetical equation(s) must equal zero...

x-11=0, x=11

x+3=0, x=3

 

Let's return to the original parabolic equation...

y^2 = 2x^2 - 17

x=3

y²=[(2)(3²)]-17

y²=18-17

y²=1, y=+ or - 1

Let's return to the original linear equation...

y=x-4

y=3-4

y=-1

 

(3,-1)

 

Original parabolic equation...

y^2 = 2x^2 - 17

x=11

y²=[(2)(11²)]-17

y²=[(2)(121)]-17

y²=242-17

y²=225

y=+ or - 15

Original linear equation...

y=x-4, y=15-4, y=11

(11,15)

 

 

y^2 = 2x^2 - 17

y=11