Nature of roots of quadratic equations

In summary, the equation kx2 - 3x + (k+2) = 0 has two distinct real roots. You can find the set of possible values of k by solving the equation for k. The Attempt at a Solution provides the b2-4ac>0 and 9-4(k+2)(k)>0. The final solution is -4k2-8k+9>0.
  • #1
thornluke
37
0

Homework Statement


The equation kx2 - 3x + (k+2) = 0 has two distinct real roots. Find the set of possible values of k.


Homework Equations


Since the equation has two distinct real roots, b2 - 4ac > 0

The Attempt at a Solution


b2-4ac>0
9-4(k+2)(k)>0
9-4(k2+2k) >0
9-4k2-8k>0
= -4k2-8k+9>0
Multiply both sides by -1,
4k2+8k-9>0
(4k-3)(k+3)>0

-3<k<3/4

However the answer is -2.46<k<0.458
I'm lost, help please! :confused:
 
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  • #2
thornluke said:

Homework Statement


The equation kx2 - 3x + (k+2) = 0 has two distinct real roots. Find the set of possible values of k.


Homework Equations


Since the equation has two distinct real roots, b2 - 4ac > 0

The Attempt at a Solution


b2-4ac>0
9-4(k+2)(k)>0
9-4(k2+2k) >0
9-4k2-8k>0
= -4k2-8k+9>0
Multiply both sides by -1,
4k2+8k-9>0
(4k-3)(k+3)>0

-3<k<3/4

However the answer is -2.46<k<0.458
I'm lost, help please! :confused:
Your factoring is incorrect.

(4k-3)(k+3) = 4k2 + 9k - 9 .

Also, multiplying by -1 will change > to < .

Solve 4k2+8k-9 = 0 by using the quadratic formula --- or by completing the square.
 
  • #3
SammyS said:
Your factoring is incorrect.

(4k-3)(k+3) = 4k2 + 9k - 9 .

Also, multiplying by -1 will change > to < .

Solve 4k2+8k-9 = 0 by using the quadratic formula --- or by completing the square.

((-8 ± √208)/8) < 0

-2.80 < k < 0.80

I'm getting closer to the "answer" (-2.46<k<0.458) am I wrong, or is the "answer" wrong?
 
  • #4
I'm getting the same roots (-2.80 and 0.80).

When I played around with the coefficients of the original quadratic, I found that if you made the coefficient of the x2 term 2k:
2kx2 - 3x + (k+2) = 0
You will get the original answer that you stated: -2.4577 < k < 0.4577. So it looks like either you copied the problem incorrectly or the book has a typo somewhere.
 
  • #5
The textbook's answer is consistent with -8k2 - 16k +9 > 0 . equivalent to -4k2 - 8k + 9/2 >0

It's hard to see how that's from a simple Typo -- unless the coefficient of x is should have been 3/√2 in the initial equation.
 
  • #6
eumyang said:
I'm getting the same roots (-2.80 and 0.80).

When I played around with the coefficients of the original quadratic, I found that if you made the coefficient of the x2 term 2k:
2kx2 - 3x + (k+2) = 0
You will get the original answer that you stated: -2.4577 < k < 0.4577. So it looks like either you copied the problem incorrectly or the book has a typo somewhere.

I guess everyone makes mistakes.. it gets extremely annoying when textbooks provide you with the wrong answers or have a typo.
 
  • #7
thornluke said:
I guess everyone makes mistakes.. it gets extremely annoying when textbooks provide you with the wrong answers or have a typo.

They are put there to help you to build self-confidence. :biggrin:
 

FAQ: Nature of roots of quadratic equations

What is the nature of the roots of a quadratic equation?

The nature of the roots of a quadratic equation depends on the discriminant, which is the value under the radical sign in the quadratic formula. If the discriminant is positive, the equation will have two distinct real roots. If the discriminant is zero, the equation will have one real root, also known as a double root. If the discriminant is negative, the equation will have two complex conjugate roots.

How do you find the discriminant of a quadratic equation?

The discriminant of a quadratic equation is found by taking the square root of b²-4ac, where a, b, and c are the coefficients of the quadratic equation in standard form: ax²+bx+c=0. This value is then used to determine the nature of the roots.

Can a quadratic equation have no real roots?

Yes, a quadratic equation can have no real roots if the discriminant is negative. This means that the roots will be complex numbers, which are not considered real numbers.

What is the difference between real and complex roots?

Real roots are numbers that can be found on the number line, while complex roots are numbers that involve the imaginary unit, i, and cannot be found on the number line. Real roots are also considered simpler than complex roots, as they do not involve imaginary numbers.

How does the graph of a quadratic equation relate to the nature of its roots?

The graph of a quadratic equation is a parabola, which can either intersect the x-axis at two distinct points, one point (when the parabola is tangent to the x-axis), or no points (when the parabola does not intersect the x-axis). These correspond to the nature of the roots being two distinct real roots, one real root, or no real roots, respectively.

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