Quadratic simultaneous eqns! Please help!?

Question

Solve the simultaneous eqns below by showing me the steps clearly..very clearly so that i can understand it properly. Thanks a lot in advance.

y = 4x2 + 3ax - 2a2
y = 2x2 + 2ax - a2 ; where a is a real number.

Note that 4x2 is 4x squared
2a2 is 2a squared
2x2 is 2x squared
a2 is a squared

And by the way, the solution for this question says: (-a, -a2) (a/2 , a2/2). However, I dont knw how the the hell its been obtained. I guess that is why im posting the question here..!!

Thanks once again!

Answer 1

Because both equations have y at the left side, the two right sides must be equal to each other:
4 x^2 + 3 a x - 2 a^2 = 2 x^2 + 2 a x - a^2

Subtraction the right side from the left, we find
2 x^2 + a x - a^2 = 0

which is a quadratic equation in one variable. It can be factored:
(2 x - a) (x + a) = 0

That means that either
2x - a = 0 ... or ... x + a = 0
so
x = a/2 ... or ... x = -a

The corresponding values for y are found by substituting these results in one of the original equations:
y = 4(a/2)^2 + 3a(a/2) - 2a^2
... = a^2 + (3/2) a^2 - 2 a^2 = a^2 / 2

y = 4(-a)^2 + 3a(-a) - 2(-a)^2
... = 4a^2 - 3a^2 - 2a^2 = -a^2

Answer 2

At an intersection point, the y-values must be equal, so

4x^2 + 3ax - 2a^2 = 2x^2 + 2ax - a^2

Rearrange:

2x^2 + ax - a^2 = 0

To find the x-values that satisfy this, use the quadratic formula:
x = (-a +/- sqrt(a^2 - 8(-a^2)))/4
= (-a +/- sqrt(9a^2))/4
= (-a +/- 3a)/4
= -a or a/2.

Now plug both of these x-values into either of the original equations to get the corresponding y-values.

Answer 3

Fortunately it is very easy as both are y

equating the 2
4x^2 + 3ax - 2a^2 = 2x^2+2ax - a^2
this reduces to a quadratic equation
2x^2 + ax -a^2 =0
factor
2x^2 + 2ax - ax - a^2 =0
or 2x(x+a)-a(x+a) = 0
(2x-a)(x+a) = 0
x= a/2 or -a
for x = a/2 calulate y by putting in either form that is a^2/2
for x = -a y = -a^2

Answer 4

y = 4x^2 + 3ax - 2a^2
y = 2x^2 + 2ax - a^2

Set the two expressions for y equal to each other.

4x^2 + 3ax - 2a^2 = 2x^2 + 2ax - a^2
2x^2 + ax -a^2 = 0

math_kp cleverly replaced + ax with + 2ax - ax to produce an expression that can be factored by grouping. If you didn't see that, you could always factor the expression as is

(2x - a)(x + a)

Answer 5

assuming equations are 2x^2 +4x - 7 = 0 and x = (-4 +/- ?p) / q Take the 1st equation and decide it utilising quadratic formula: a = 2, b = 4, c = -7: x = (-b +/- ?(b^2-4ac)) / 2a x = (-(4) +/- ?((4)^2 -4(2)(-7))) / 2(-7) x = (-4 +/- ?(sixteen -(-fifty six))) / -14 x = (-4 +/- ?seventy 2) / -14 it rather is corresponding to the various EQUATION.... p = seventy 2, q = -14

Answer 6

y=4x^2+3ax-2a^2
2y=4x^2+4ax-2a^2
y-ax=0
y=ax
substituting
ax=2x^2+2ax-a^2
2x^2+ax-a^2=0
x=[-a+/-(a^2+8a^2)]/4
x=-a(1+/-3)/4=a/2 or -a
y=a^2/2 os -a^2