If one root of the equation 2x^2 + 3x – k = 0 is 6, what is the value

We just enter this root into the equation in order to recieve an equation to find the answer!

2*6^2+3*6-k=0
k=72+18=90

The answer is A

The question should say that k is a constant. In that case, if 6 is a 'root' of the equation, that just means that x=6 is one solution to the equation. So we can plug in x=6:
\(\begin{align*}\\
2x^2 + 3x - k &= 0 \\\\
2(6^2) + (3)(6) - k &= 0 \\\\
72 + 18 &= k \\\\
90 &= k\\
\end{align*}\)

let the other root is 'a'
a +6=-3/2; a = -15/2
6a = -k/2
implies k=90

Substituting 6 for x, we have:

2(36) + 3(6) - k = 0

72 + 18 = k

90 = k

Answer: A

When we have a value of one root in a quadratic equation, make the RHS to zero (0) so that we can substitute the given root to find the other variables.

In this case, the RHS is already 0. So just substitute the given root value for x, i.e, x=6

2*\((6)^2\) + 3*(6) - k = 0

2*(36) + 18 = k

72 + 18 = k

90 = k

Therefore the answer is A

Given: one root of the equation 2x^2 + 3x – k = 0 is 6

Asked: what is the value of k?

Putting x=6

2*6^2 +3*6 =k
k= 72+18=90

IMO A

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If one root of the equation 2x^2 + 3x – k = 0 is 6, what is the value of k?

Putting x=6, we get,
2(6^2) + 3(6) - k = 0
or,2(36) + 3(6) - k = 0
or,72 + 18 = k
or,k=90

correct answer A

IanStewart ScottTargetTestPrep

Thanks for your solutions - very helpful. I do have a follow up question though.

If k is 90, isn't the other root -15? In which case, how do we end up with a 3 for the middle term?

My work below:
2x^2 + 3x - 90 = 0
(2x - 6) (2x + 15) = 0

Adding -6 and 15 gives me 9, not 3. Really don't know where I'm going wrong

If 6 is a root of a quadratic, then x-6 is a factor of the quadratic (not 2x-6). In order to get 2x^2 when we multiply our two factors together, the other factor will need to begin with "2x", and in order to get -90 at the end when we multiply out, the other factor will need to end in "+15", so the factorization will be

(x-6)(2x + 15)

and if you multiply the x term from each by the number in the other factor, you'll get -12x + 15x = 3x.

I think you are referring to the rule that the constant term in a quadratic equation in the standard form is the product of the roots, but remember that it only works for quadratics of the form x^2 + bx + c = 0, where the coefficient of the x^2 term is 1.

For a general quadratic of the form ax^2 + bx + c = 0, the product of the roots is c/a, so the product of the roots of the equation 2x^2 + 3x - 90 = 0 is c/a = (-90)/2 = -45, which means that the other root of the equation is -45/6 = -15/2. So you can rewrite the quadratic in the factored form as a(x - p)(x - q) = 2(x - 6)(x + 15/2) = (x - 6)(2x + 15). Also, the sum of the roots of an equation of the form ax^2 + bx + c = 0 is -b/a, which means the sum of the roots of 2x^2 + 3x - 90 = 0 is -3/2. We can verify that the sum of the roots is indeed 6 + (-15/2) = -3/2.

 

\(­2x^2 + 3x – k = 0\)

Or, \(2*6^2 + 3*6 – k = 0\)

Or, \(72 + 18 - k =0\)

So, \(k = 90\), Answer will be (A)