We know that in a Quadratic Equation ax^2 + bx + c = 0,
Sum of roots = -b/a
Product of roots = c/a
Here, we have the equation 8x^2 - 6x + p = 0.
It is given that one root of this equation is square of the other root. So, if we assume one root to be 'k', the other root can be 'k^2'. So, we assume the roots to be k and k^2.
Using the product of roots formula,
k * k^2 = p/8
=> k^3 = p/8 (equation 1)
Using the sum of roots formula,
=> k + k^2 = 3/4
=> 4k + 4k^2 = 3
=> 4k^2 + 4k - 3 = 0
Here, we get a basic idea about how we are going to solve the problem. Applying the information given in the question, we get another Quadratic Equation in k which can be easily solved and value (or maybe values) of k can be obtained. Using those values of k in equation 1, p can be determined. In the process, we should not forget what is k. k is an assumption we made for a root of the equation given in the question.
This one is an easy one and can be solved by 'splitting the middle term'. Had it been a difficult one, the Quadratic Formula could have been used.
=> 4k^2 + 4k - 3 = 0
=> 4k^2 - 2k + 6k - 3 = 0
=> 2k(2k - 1) + 3(2k - 1) = 0
=> (2k + 3)(2k - 1) = 0
=> 2k + 3 = 0 ; 2k - 1 = 0
=> k = -3/2; k = 1/2
We'll now use these values of k in equation 1 one by one to obtain the value of p.
Using k = -3/2,
k^3 = p/8
=> (-3/2)^3 = p/8
=> -27/8 = p/8
=> p = -27
Using k = 1/2
k^3 = p/8
=> (1/2)^3 = p/8
=> 1/8 = p/8
=> p = 1
Hence the possible values of p are 1 and -27.
P.S. - To cross-check the solution, use the values of p from the answer in the equation given in the question. You'll get the equations 8x^2 - 6x + 1 = 0 and 8x^2 - 6x - 27 = 0. If you solve these equations, you'll see that in both of them, one root is square of the other. That verifies that the answer is correct.
Thankyou so much ....this helped me a lot ....love u
ReplyDeleteGlad to hear such kind words. You're welcome!
DeleteIf one root of a quadratic equation ax^2+bx+c is square the other then prove b^3+ac^2+a^2c=3ab
ReplyDeleteAlright mate. First off, there is an error in the question you've asked. Please do me and yourself a favor and pay attention to when you're doing maths. That's what it is all about.
DeleteIt should be b^3 + ac^2 + a^2c = 3ABC. Take the quadratic equation x^2 - 6x + 8 = 0 (Roots 2, 4) and verify it yourself)
Now here's the solution. The equation is ax^2 + bx + c = 0. Assume the roots to be t and t^2. Now, make the equations for sum and products of roots.
t + t^2 = -b/a eq.1
t^3 = c/a eq.2
Take t = (c/a)^1/3 from eq.2 and substitute into eq.1
(c/a)^1/3 + (c/a)^2/3 = -b/a
Cubing both the sides of the equation
=> [(c/a)^1/3 + (c/a)^2/3 ]^3 = (-b/a)^3
=> (c/a) + (c/a)^2 + 3 . (c/a)^1/3 . (c/a)^2/3 . [(c/a)^1/3 + (c/a)^2/3 ]= -b^3/a^3
Look at the third term in LHS. The (c/a)^1/3 + (c/a)^2/3 part of it is simply what we got after substituting eq 1 and 2, which is b/a
=> (c/a) + (c/a)^2 + 3 . (c/a)^1/3 . (c/a)^2/3 . (-b/a) = -b^3/a^3
=> c/a + c^2/a^2 - 3bc/a^2 = -b^3/a^3
=> (ac + c^2 -3bc)/a^2 = -b^3/a^3
=> ac + c^2 -3bc = -b^3/a
=> a^2.c + a.c^2 - 3abc = -b^3
=> b^3 + a^2.c + a.c^2 = 3abc
If one root of a quadratic equation ax^2+bx+c is square the other then prove b^3+ac^2+a^2c=3ab
ReplyDeleteAlright mate. First off, there is an error in the question you've asked. Please do me and yourself a favor and pay attention to when you're doing maths. That's what it is all about.
DeleteIt should be b^3 + ac^2 + a^2c = 3ABC. Take the quadratic equation x^2 - 6x + 8 = 0 (Roots 2, 4) and verify it yourself)
Now here's the solution. The equation is ax^2 + bx + c = 0. Assume the roots to be t and t^2. Now, make the equations for sum and products of roots.
t + t^2 = -b/a eq.1
t^3 = c/a eq.2
Take t = (c/a)^1/3 from eq.2 and substitute into eq.1
(c/a)^1/3 + (c/a)^2/3 = -b/a
Cubing both the sides of the equation
=> [(c/a)^1/3 + (c/a)^2/3 ]^3 = (-b/a)^3
=> (c/a) + (c/a)^2 + 3 . (c/a)^1/3 . (c/a)^2/3 . [(c/a)^1/3 + (c/a)^2/3 ]= -b^3/a^3
Look at the third term in LHS. The (c/a)^1/3 + (c/a)^2/3 part of it is simply what we got after substituting eq 1 and 2, which is b/a
=> (c/a) + (c/a)^2 + 3 . (c/a)^1/3 . (c/a)^2/3 . (-b/a) = -b^3/a^3
=> c/a + c^2/a^2 - 3bc/a^2 = -b^3/a^3
=> (ac + c^2 -3bc)/a^2 = -b^3/a^3
=> ac + c^2 -3bc = -b^3/a
=> a^2.c + a.c^2 - 3abc = -b^3
=> b^3 + a^2.c + a.c^2 = 3abc
Find the value of k, for which one root of the quadratic equation kx2
ReplyDelete-14x+8=0 is
six times the other.
Let the roots be a and 6a.
ReplyDeleteSum of roots = 7a
=> 14/k = 7a
=> ak = 2
Product of roots = 8/k
=> 6a^2 = 8/k
=> 3a^2 = 4/k
=> 3a^2.k = 4
=> 3a.ak = 4
=> 3a(2) = 4 (we found ak = 2 above)
=> a = 4/6 = 2/3
Substituting this value of a in ak = 2
=> 2k/3 = 2
=> k = 3
k = 3 is the answer.
Roots of the quadratic equation x
ReplyDelete2 + px + q = 0 are 2 and 15. Determine its roots if the value of q is doubled. pls solv
Thank you very much you are great.
ReplyDeleteHello, can someone solve and explain...8x^2+px+1=0...find p if one of the roots of this equation is the square of the other... Thank you
ReplyDeleteIf one of the roots of equation 8X^2+px+1=0 is the square of the other ,find p
ReplyDeleteIt's too bigger bro, need some shortcut for fill up mcq in 30 sec. You know, 2 lakh students are participate in University admission exam for only 2000 seats. So its very difficult when you have to fill up 100 mcq in 60 minutes.
ReplyDelete