Find the set of values of k for which the equation 2x^2+kx+2+0 has no real roots ?

Question

Find the set of values of k for which the equation 2x^2+kx+2+0 has no real roots ?

Once again i have become stuck on another maths equations.
Please can someone give me some guidence on this question.
Thank you for any help.

Answer 1

Assuming the equation is

2x^2+kx+2=0

Using the quadratic function you find the roots are

x = [-k ± √(k² - 16)]/4

The roots are real only when k²≥16 or |k|≥4

So as long as |k|<4 there are no real roots. The interval for k is then

-4

Answer 2

If the quadratic equation ax^2+bx+c = 0 has no real roots, then b^2 - 4ac < 0 With 7x^2 + kx + 8 = 0, b^2 - 4ac = k^2 - 4(7)(8) = k^2 - 224 Hence for no real roots, k^2 - 224 < 0 i.e. -4√14 < k < 4√14 or k lies in the interval (-4√14, 4√14)

Answer 3

Find the set of values of k for which the equation 2x^2+kx+2+0 has no real roots ?

When using the quadratic formula to solve a quadratic equation ax^2 + bx + c = 0, the discriminant is b^2 - 4ac. This discriminant can be positive, zero, or negative. (When the discriminate is negative, then we have the square root of a negative number. This is called an imaginary number, sqrt(-1) = i. )

For a = 2, b = k and c =2, b^2 - 4ac = k^2 – 16

If this is < 0, then k^2 < 16 and k < 4

But k > -4 also satisfies the inequality, hence

-4 < k < 4

Answer 4

In general, the roots of ax^2+bx+c=0 are given by the formula:

x = ( -b "+/-" (b^2 - 4 a c)^{1/2} ) / (2 a)

where "+/-" means it can be a + there or a -... hence usually you will have two roots. Observe that both roots will be real if b^2-4ac >= 0, and both roots will be non real (complex) if b^2-4ac<0.

In your particular case, ax^2+bx+c=2k^2+kx+2, so a=2, b=k and c=2. Then the roots of your equation will be not real if b^2-4ac=k^2-4*2*2<0, which means that k^2<16. Taking square roots, you have (| . | means absolute value) | k | < 4. (why?)

Answer 5

k^2 < 4

Answer 6

you look at the determinant b^2-4ac
det = k^2-16 = (k-4)(k+4)
to have no real roots this expression must be negative

it is negative for values of k

-2 all values ok between -2 and+2 with the limits k=2 or k =-2 not incluted

Answer 7

2x^2+kx+2=0
discriminanat=k^2-16.
there will beno real roots, if discrimi <0
k^2-16<0
k^2<16
-4

Answer 8

b^2-4ac<0
a=2,b=k,c=2
k^2-16<0
k^2<16
-4