Find all values of k such that the equation has exactly one root?

Question

x^2+kx+25=0

the answer is {10, -10}

Can someone tell me how to get there? Thanks.

Answer 1

if the eqn has exactly one root that means roots are equal, you need to use delta (b^2-4ac). for eql roots it will =0.

so b^2-4ac = 0
k^2 - 100 = 0
(k + 10)(k - 10) = 0
k = +- 10

Answer 2

In order for a quadratic equation to have exactly one root, you must solve for the discriminant. To do so, note that the form of a quadratic equation is as follows:

ax^2 + bx + c = 0

The discriminant is calculated by the formula b^2 - 4ac.

Note: Recall that the quadratic formula goes like this

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

There are times when this would yield two solutions or even zero real solutions, but there's even a special time when it would yield exactly one solution. If I had an answer of 3 "plus or minus" 5, then the answer would be 3+5 and 3-5, or 8 and -2.

However, what if I had something like 9 "plus or minus" 0? I'd get two of the same answer, because "plus or minus 0" doesn't really give two separate answers.

That's EXACTLY why you require b^2 - 4ac to be 0 in order to have exactly one root. Let's calculate b^2 - 4ac.

In our case, a = 1 (the coefficient of x^2), b = k (the coefficient of x), and c = 25 (the constant). Thus

b^2 - 4ac = k^2 - 4(1)(25)
or
k^2 - 100

Now, we WANT this to equal to 0, because that's how we get exactly one solution

k^2 - 100 = 0

And now we solve this as normal.

First, factor it as a difference of squares:

(k - 10) (k + 10) = 0
k = 10 or k = -10

Those are our values for k. What they are saying is:

x^2 + 10x + 25 = 0 and
x^2 - 10x + 25 = 0

have EXACTLY one root.

It's no surprise, because both of them are perfect squares, which factor into

(x+5)^2 = 0 and
(x - 5)^2 = 0

Answer 3

Use the discriminant: b^2 - 4ac. in case you have an equation interior this style of ax^2 + bx + c = 0, it could have 2 unique recommendations if the discriminant is extra effective than 0, one answer if the discriminant is comparable to 0, and no recommendations if that's decrease than 0. So interior the case of x^2 + kx + seventy 9=0, we've a=a million, b=ok, and c=seventy 9. Plug this into the discriminant and you get ok^2 - 4(a million)(seventy 9), or ok^2 - 316. So for the unique equation to in straightforward terms have one root, this might desire to equivalent 0. that's genuine whilst ok = sqrt(316) or -sqrt(316). the clarification the discriminant works turns into sparkling in case you check out the quadratic formula. The discriminant is the section discovered below the novel. in case you may take the sq. root, then the "plus or minus" interior the quadratic formula provides 2 solutions. If the discriminant is 0 then the sq. root of it relatively is 0, and the "plus or minus" 0 interior the quadratic formula in straightforward terms provides the comparable answer two times. If the discriminant is decrease than 0, then you truly have not have been given any answer, when you consider which you may no longer take the sq. root of a adverse style.

Answer 4

An equation has exactly one square root if and only if its discriminant is zero.

The discriminant is the part underneath the √ sign in the quadratic equation.

b² - 4ac = 0
k² - 4(1)(25) = 0
k² - 100 = 0
k² = 100
k = +/- 10

Check:

x² - 10x + 25 = 0
(x - 5)² = 0
x = 5

x² + 10x + 25 = 0
(x + 5)² = 0
x = -5

Answer 5

x²+kx+25=0

For the equation has exactly one root:
Delta = 0
=>
b²-4ac = 0
k² - 4*1*25 = 0
k² = 100

k = 10

₢

Answer 6

ax^2+2xab+b^2=(ax+b)^2 or ax^2-2xab+b^2=(ax-b)^2

consider the case 1: ax^2+2xab+b^2=(ax+b)^2

here x^2+kx+25=0

a=1
b^2=25 ==> b=5
2xab=kx ==> 2ab=k ==>2*1*5=k ==>k=10

therefore,
ax^2+2xab+b^2=(ax+b)^2

1*x^2+2x*1*5+5^2=(1x+5)^2 ==> (x+5)^2=0 ==>x=-5

now consider case 2: ax^2-2xab+b^2=(ax-b)^2

x^2+kx+25=0

ax=x ==>a=1
b^2=25 ==> b=5
-2xab=kx ==> -2ab=k ==> -2*1*5=k ==>k=-10

therefore,
ax^2-2xab+b^2=(ax-b)^2

1*x^2-2x*1*5+5^2=(1x-5)^2 ==> (x-5)^2=0 ==>x=5

Answer 7

look for two numbers whose product is 25 and the sum is k
5 and 5
-5 and -5