Quadratic Equations!?

Question

Question 1.

The roots of the equation
x square + ( p -10 )x = 10p
are alpha and alpha+4. Find the possible values of p.

Question 2:
The roots of the equation
x square - 8x +h = 0
are alpha and alpha+3k. Express h in terms of k.


Can anyone help me with this 2 questions?

Answer 1

1)Add 2 roots(L=alpha)
L+L+4=2L+4

2)multiply 2 roots
L(L+4)=L^2+4L

3)write the equation{etc:x^2-(add 2 roots)x+(multiply2 roots)}
x^2-(2L+4)x+L^2+4L

4)compared the equation to X^2+(p-10)x=10p
i)2L+4=-(p-10)
2L+4=-p+10
2L+p=6
p=6-2L
replace p=6-2L into equation(ii)L^2+4L=-10p
L^2+4L=-10(6-2L)
L^2+4L=-60+20L
L^2-16L+60=0
(L-10)(L-6) =0(factorise)
L=10 or L=6
When L=10,p=6-2(10)
=6-20
=-14
When L=6,p=6-2(6)
=6-12
=-6
So,p=-14 and p=-6
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1)Add 2 roots:
L+L+3k=2L+3k

2)multiply 2 roots:
L(L+3k)=L^2+3Lk

3)write equation:
x^2+(2L+3k)x+L^2+3Lk

4)compared to x^2-8x+h=0

5)(i)2L+3k=-8
(ii)L^2+3Lk=h

6)From( i)2L+3K=-8
2L=-8-3k
L=-4-3/2k

7)replace L=-4-3/2k into (ii)
(-4-3/2k )^2+3(-4-3/2k)k=h
16+12K+9/4k^2-12-9/2k =h
4+15/2k+9/4k^2 =h

8)So,h=9/4k^2+15/2k+4

Answer 2

1)

x^2 + (p - 10)x = 10p

First, move everything to the left hand side.

x^2 + (p - 10)x - 10p = 0

If a and (a + 4) are the roots, it follows that (x - a) and (x - (a + 4)) are factors. Multiplying them together, we have

(x - a)(x - (a + 4)) = x^2 - x(a + 4) - ax + a(a + 4)

Simplifying this by grouping like terms, we have

x^2 - x (a + 4 + a) + a(a + 4)
x^2 - (2a + 4)x + a(a + 4)

Now, all we have to do is equate this to the earlier equation:

x^2 + (p - 10)x - 10p = x^2 + (-2a - 4)x + a(a + 4)

At this point, we can equate them component-wise. That is, each coefficient on the left hand side is going to equal the coefficients on the right hand side.

p - 10 = -2a - 4
-10p = a(a + 4)

We're interested in solving for the possible values for p, and this is just a 2-equations, 2-unknowns problem. We're going to use substitution by solving for p in the first equation and plugging it into the second.
p - 10 = -2a - 4 means p - 6 = -2a, so (-1/2)(p - 6) = a

Plugging this into the second equation, -10p = a(a + 4), we have

-10p = [(-1/2)(p - 6)] [(-1/2)(p - 6) + 4]

All we have to do is solve for p.

-10p = [-1/2] (p - 6) ( (-1/2)(p - 6) + 8/2]
-10p = [-1/2] (p - 6) ( (-p + 6 + 8)/2 )
-10p = [-1/2] (p - 6) ( (-p + 14)/2 )
-10p = [-1/2] (p - 6) ( (-1/2) (p - 14) )
-10p = [1/4] (p - 6) (p - 14)

Multiply both sides by 4, to get

-40p = (p - 6) (p - 14)

Now, FOIL out the right hand side.

-40p = p^2 - 20p + 84

Move the -40p to the right hand side, to get

0 = p^2 + 20p + 84

This factors into

0 = (p + 14) (p + 6)

Therefore, our possible values for p are
p = {-14, -6}

2) x^2 - 8x + h = 0

The roots are given to be a and (a + 3k). This means the corresponding factors will be
(x - a) (x - (a + 3k))

FOILING this out, we get.

x^2 - x(a + 3k) - ax + a(a + 3k)

And now, combining the x terms, we get

x^2 + (-a - 3k - a)x + a(a + 3k)
x^2 + (-2a - 3k)x + a(a + 3k)

Equating this to the existing quadratic, we have

x^2 - 8x + h = x^2 + (-2a - 3k)x + a(a + 3k)

And now we equate component-wise. Remember our goal is to solve for h in terms of k.

-2a - 3k = -8
a(a + 3k) = h

We're going to solve for "a" in the first equation, and then substitute that value into the second equation.

-2a - 3k = -8
2a + 3k = 8
2a = -3k + 8, so a = (1/2)(-3k + 8)

Plugging this into a(a + 3k) = h, we get

(1/2) (-3k + 8) [ (1/2) (-3k + 8) + 3k ] = h

Let's multiply out that (1/2) by (-3k + 8), as we WANT a common denominator in there.

(1/2) (-3k + 8) [ [-3k + 8]/2 + 6k/2 ] = h
(1/2) (-3k + 8) [ [-3k + 8 + 6k]/2 ] = h

Now, we can pull out the (1/2), merge it with the (1/2) on the outside of all the brackets, to get 1/4.

(1/4) (-3k + 8) (3k + 8) = h

Note that the bracketed terms are now conjugates of each other. If it's not obvious, let's swap the terms like this.

(1/4) (8 - 3k) (8 + 3k) = h

It FOILS as a difference of squares.

(1/4)(64 - 9k^2) = h

And that is h in terms of k, so that's our final answer.

Answer 3

alpha*(alpha+4)= -10p(product) of the roots

2*alpha+4=10-p (sum of bthe roots) ==> alpha=(6-p)/2

going to the first (6-p)/2*( (6-p)/2+4)= - 10p ypu get operating

p^2+20p+84=0 p=-14 p=-6



alfa*(alfa+3k)=h and 2alfa +3k=8 ==> alfa = (8-3k)/2

(8-3k)/2 * ( (8-3k)/2 +3k) =h ..Youshould make the indicated oper.

Answer 4

1.)
let the roots be A,B
A-B=+or-4 or (A-B)^2=16
(A-B)^2=(A+B)^2-4*A*B
sum of roots = -coeff of x/ coeff of x^2
product of roots = constant term/ coeff of x^2
(A-B)^2 =(p-10)^2 + 40*p
16 = p^2 + 20p +100
(p+10)^2=16
p+10=+or-4
p= -14 or p= -6
Try to solve the 2nd one using this rationale.