HELP, guys!!!!!?

Question

1. Find the value of a such that the line y = 2x + a has exactly one intersection point with the parabola with equation y = x squared + 3x + 2.

Please answer this question by giving me each n every step because i have got to understand the logic behind it rite? But, just so that you knw, the correct answer is 1.75 and i dont knw how that answer was obtained.

Thanks in advance, guys!!

Answer 1

You want to find the one point that will satisfy both equations...

You have two equations both in forms of y:
y = 2x + a
y = x² + 3x + 2

So equate the two formulas:
2x + a = x² + 3x + 2

Now put everything on one side:
x² + 3x + 2 - 2x - a = 0

And combine coefficients on the like terms:
x² + x + (2 - a) = 0

You want this to have *one* solution.

Using the quadratic formula:
ax² + bx + c = 0

x = [ -b ± √(b² - 4ac) ] / 2a

Notice how the only thing that matters is the item under the square root sign. This is called the "discriminant". If it is 0, then you can see that you get ±√0 which only results in one value for x.

So just take your values from before and plug them into the discriminant.
x² + x + (2 - a) = 0

Your coefficients are:
1
1
(2-a)

Plugging this into the discriminant:
b² - 4ac = 0
1² - 4(1)(2 - a) = 0
1 - 4(2 - a) = 0
1 - 8 + 4a = 0
-7 + 4a = 0
4a = 7
a = 7/4

You can turn this into a decimal (1.75) which is the same as your supplied result.

So there is your answer. The value of a = 7/4 will make the line intersect the parabola at a single point.

Note: you can actually continue figuring the value of x
x = ( -1 ± 0 ) / 2
x = -1 / 2

Now plug that into one of the original equations to figure y:
y = 2x + 7/4
y = 2(-1/2) + 7/4
y = -1 + 7/4
y = 3 / 4

So when you set a = 7/4, then you'll have an intersection point at (-1/2, 3/4). I know this is more than you asked, but I thought you might like to see the full solution. The first link below is a picture showing the graph of the parabola and the line...

Answer 2

first,all other guys answered right but i will repeat their answer and add another
u equalize them :
x ^ 2 + 3 x +2 = 2x + a
x ^ 2 + x + (2 - a) = 0
use the perfect square method
x^2 + x + 0.25 + (1.75 - a) = 0
(x+0.5)^2 + (1.75 - a) = 0
(x+0.5)^2 = a - 1.75
x +0.5 = +ve or -ve square root of (a-1.75)
for the right hand side to have one value it should equal zero otherwise it may be +ve or -ve
Then
a - 1.75 = 0
a = 1.75

The other solution
the line touches the curve in one point , so it is a tangent to it
and if u r familiar with calculus u will understand this
if the line y=2x + 3 is a tangent to the cuve y=x^2+3x+2
then the co efficient of x in the line ( 2 ) can be obtained from the first derevative
f'(x)=2x + 3
then 2x + 3 = 2
2x = -1
x = -0.5
i.e.at the point (-0.5,y) the two graphs touch
y of this point can be obtained from the curve's equation
y= (-0.5)^2 + 3 *-0.5 + 2
=0.25 -1.5 + 2
=0.75
since the two lines touch the two "y"s are the same
then y=2x +a
0.75 = 2*-0.5 + a
0.75 = -1 + a
a = 1.75

I am sorry this answer is too long but i thought it might help
if u need any help with this solution or any other problem plzcontact me at [e-mail]

Answer 3

Since the both equations are same
x^2 + 3x + 2 = 2x + a

Form a quadratic equation:
x^2 + x + 2 - a = 0

or 1 - 4 (2-a)=0
or 1 = 8-4a
or 4a= 7
or a = 7/4
or a = 1.75

Answer 4

Thanks for including the answer so that I know my own steps are correct.

Step 1: set the two equations equal to each other:
2x+a = x^2+3x+2

Step 2: Get the equation equal to zero.
x^2+1x+(2-a) = 0

Step 3: Complete the square
x^2+1x +1/4 = -2 +a + 1/4
(x+1/2)^2 = -1.75 + a

Step 4: Since you want your perfect square to equal exactly to zero you need to set -1.75+a = 0 and solve

So...
a = 1.75 to make sure that your line only hits the parabola one time.

Hope this helps. Good Luck.

Answer 5

Set the y-values equal to each other:

2x + a = x^2 + 3x + 2.

Rearrange so everything is on one side:

x^2 - x + (2-a) = 0.

Obtain the roots, using the quadratic formula:

x = (1 +/- sqrt(1 - 4(2-a))) / 2.

Focus on this part: sqrt(1-4(2-a)) = sqrt(4a-7).

If 4a-7 > 0, then there are two real roots of this equation, so there are two intersections.

If 4a-7 < 0, then there are no real roots, so no intersections.

If 4a-7 = 0, then there is one real root (a double root), so there is one intersection.

Solve this equation for a and you get the value indicated in the question.

Answer 6

Set them equal to each other:

x^2 + 3x + 2 = 2x + a

Form a quadratic equation:

x^2 + x + 2 - a = 0

Set the discriminat equal to zero:

1 - 4(2-a)=0

Solve:

1 = 8-4a

4a = 7

a = 7/4

Answer 7

SEE SUBSTITUTE THE VALUES..
LET SAY IN THE equataion of parabola[ y=x^2+3x+2 ]
put the value of x=(y-a)/2 in y=x^2+3x+2
u will be getting a quadratic equation of y variabel ,solve it , the answer wikll be two, one negative and an other positive. both will be correct .
the answer will be y
than u put the value of y in equaion x=(y-a)/2 u will again get to answers wd respect to the two values of 'y'..
and the answer will always come in 'x,y'
best of luck!