Verified answer

there is a sub part of quadratic formula called the discriminate that given ax^2 +bx +c the discriminate is b^2 - 4ac

in the quadratic formula the you take the square root of the discriminate

this results in:

* if discriminate is greater than zero there will be two distinct real roots,

* if the discriminate equals zero then the will be two roots but they have the same value. (so you effectively only get one answer).

* if the discriminate is less than zero this results in a square root of a negative number, which mean there are no real roots, how there are two imaginary or complex roots.

[the also the fact is if the discriminate is also a perfect square than the roots will be rational]

so x²+10x+36=0, a = 1, b = 10 and c = 36

therefore the discriminate = 10² - 4*1*36 = 100 - 144 = -44 so no real roots.

if you graph the equation you will see the graph does not cross the x-axis.

if x²+10x+k=0 has two equal roots then b² -4ac = 0

=> 10² - 4 * 1 * k = 0

=> 100 - 4k = 0

=> 4k = 100

=> k = 25

now prove it x²+10x+25=0

=> x² + 5x + 5x + 25 = 0

=> x(x +5) + 5(x + 5) = 0

=> (x +5)(x +5) = 0

If this has no real roots, the stuff under the square root symbol in the quadratic formula must be negative.

That's where the b^2 - 4ac stuff comes from.

So here, a = 1, b = 10, and c = 36

10^2 - 4*1*36 = 100 - 144 = -144 < 0 so there are no real solutions to this equation.

For the second part, for the roots to be equal, that means the two solutions for the quadratic formula must be identical.

The quadratic formula can be written two ways, as a single fraction, which is how it is usually presented in textbooks, but it can also be written like this.

-b/2a (+/-) sqrt[b^2-4ac]/2a

The only way both solutions can be equal is if the stuff after the (+/-) is 0 as this is the portion of the formula that gives the two solutions.

Since this is a fraction, how do we get a fraction equal zero? The numerator must equal zero.

So sqrt[b^2 - 4ac] = 0

Then b^2 - 4ac must also = 0

So for the second part, a = 1, b = 10, c = k

10^2 - 4*1*k = 0 {Solve for k}

100 - 4k = 0

100 = 4k

25 = k

So the equation with equal roots would be x^2 + 10x + 25 which should make some sense as this is a perfect square polynomial.

I'm not sure whether you meant exactly what you wrote, or if you intended to write this: x^2 + 10x + 36 = 0

in either case, the number under the squareroot symbol is negative, so the roots are not real.

To answer your final question, again, it is not clear whether the 10 is a coefficient of x or is a constant, i.e. 10.

No real roots means it will have in I value!! So but the equation in -b±sqrt(B*B- 4ac) / 2a.. And K=25.. 5 and 5 is the equal root for K

10^2 - 4 X 36 is negative so no real roots

(x+5)^2 = x^2 + 10x + 25, so k = 25