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505-555 Level|   Algebra|                     
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If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a 18 Sep 2012, 02:57
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If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a constant, what is the other solution?

(A) -7
(B) -4
(C) -3
(D) 1
(E) 6
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Bunuel
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If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a 18 Sep 2012, 02:57
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SOLUTION

If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a constant, what is the other solution?

(A) -7
(B) -4
(C) -3
(D) 1
(E) 6

Viete's formula for the roots \(x_1\) and \(x_2\) of equation \(ax^2+bx+c=0\): \(x_1+x_2=-\frac{b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Rewrite the equation given: \(x^2 + 3x +(k-10) =0\)
Thus according to the first property \(4+x_2=-\frac{3}{1}\) --> \(x_2=-7\). As we can see we can find the value of \(x_2\) even without finding the value of \(k\).

Answer: A.
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If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a 18 Sep 2012, 06:25
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If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a constant, what is the other solution?

(A) -7
(B) -4
(C) -3
(D) 1
(E) 6

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Since 4 is one of the solution, so it must satisfy the equation.
Putting 4 as value of x: 4^2 + 3*4 + k = 10 => k =-18
so the equations is
x^2 + 3x - 28 = 0

Solving this by factorization gives => (x-4)(x+7) = 0
So x = -7 is other solution.

Hence A
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If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a 27 Jan 2018, 08:09
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Bunuel
If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a constant, what is the other solution?

(A) -7
(B) -4
(C) -3
(D) 1
(E) 6
Show more

Let's first determine the value of k.

Since x = 4 is a solution to the equation x² + 3x + k = 10, we know that x = 4 SATISFIES the equation.
That is: 4² + 3(4) + k = 10
Evaluate to get: 16 + 12 + k = 10
Solve for k to get: k = -18

So, the ORIGINAL equation is x² + 3x + (-18) = 10
This is the same as: x² + 3x - 18 = 10
We now need to solve this equation.

First, set it equal to zero: x² + 3x - 28 = 0
Factor: (x + 7)(x - 4) = 0
So, x = -7 or x = 4

We already know that x = 4 is one solution.
So, the other solution is x = -7

Answer: A

Cheers,
Brent
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If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a 18 Sep 2012, 03:13
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Slove equation by putting x=4 as it is one of the solution.
u gets value of k=-18

now put k=-18 in the equation,u get eqn in the form of ax^2+bx+c=0. Solving it we get x=4 and x=-7

so other solution is -7.
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If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a 18 Sep 2012, 05:07
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Shortcut for the question-

The sum of roots of a equation is = -b/a = -3
One of the given roots is 4
Thus the other root has to be -7
Check = 4-7 = -3 (confirm)

Thus Answer A
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If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a 18 Sep 2012, 10:28
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The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a constant, what is the other solution?

(A) -7
(B) -4
(C) -3
(D) 1
(E) 6

Practice Questions
Question: 45
Page: 158
Difficulty: 600

GMAT Club is introducing a new project: The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

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One sol. of equation is 4. therefor putting this as X in the equation and finout the constant K.
therefore K = -28.
so, equation can be rewritten as (X-4)(X+7) = 0.

other solution is X = -7.
ANswer "A"
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If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a 01 Mar 2015, 19:52
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You don't need to solve for K, we know that 4 is a solution so we already know one of the terms (x-4)(x+......) we also have x2 + 3x + k = 10, -4 + 7=3 so (x-4)(x+7):0 hence x: -7
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If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a 28 Apr 2016, 07:28
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Bunuel
If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a constant, what is the other solution?

(A) -7
(B) -4
(C) -3
(D) 1
(E) 6

Practice Questions
Question: 45
Page: 158
Difficulty: 600
Show more

The phrase “4 is one solution of the equation” means that one value of x is 4. Thus, we first must plug 4 for x into the given equation to determine the value of k. So we have

4^2 + (3)(4) + k = 10

16 + 12 + k = 10

28 + k = 10

k = -18

Next we plug -18 into the given equation for k and then solve for x.

x^2 + 3x – 18 = 10

x^2 + 3x – 28 = 0

(x+7)(x-4) = 0

x = -7 or x = 4

Thus, -7 is the other solution. Answer A.
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If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a 09 Dec 2022, 17:00
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BrentGMATPrepNow
Bunuel
If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a constant, what is the other solution?

(A) -7
(B) -4
(C) -3
(D) 1
(E) 6

Let's first determine the value of k.

Since x = 4 is a solution to the equation x² + 3x + k = 10, we know that x = 4 SATISFIES the equation.
That is: 4² + 3(4) + k = 10
Evaluate to get: 16 + 12 + k = 10
Solve for k to get: k = -18

So, the ORIGINAL equation is x² + 3x + (-18) = 10
This is the same as: x² + 3x - 18 = 10
We now need to solve this equation.

First, set it equal to zero: x² + 3x - 28 = 0
Factor: (x + 7)(x - 4) = 0
So, x = -7 or x = 4

We already know that x = 4 is one solution.
So, the other solution is x = -7

Answer: A

Cheers,
Brent
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how do I get from x² + 3x - 28 = 0 to (x + 7)(x - 4) = 0 How do you know that x² + 3x - 28=(x + 7)(x - 4) = 0?
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If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a 08 Feb 2023, 03:55
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BrentGMATPrepNow
Bunuel
If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a constant, what is the other solution?

(A) -7
(B) -4
(C) -3
(D) 1
(E) 6

Let's first determine the value of k.

Since x = 4 is a solution to the equation x² + 3x + k = 10, we know that x = 4 SATISFIES the equation.
That is: 4² + 3(4) + k = 10
Evaluate to get: 16 + 12 + k = 10
Solve for k to get: k = -18

So, the ORIGINAL equation is x² + 3x + (-18) = 10
This is the same as: x² + 3x - 18 = 10
We now need to solve this equation.

First, set it equal to zero: x² + 3x - 28 = 0
Factor: (x + 7)(x - 4) = 0
So, x = -7 or x = 4

We already know that x = 4 is one solution.
So, the other solution is x = -7

Answer: A

Cheers,
Brent


how do I get from x² + 3x - 28 = 0 to (x + 7)(x - 4) = 0 How do you know that x² + 3x - 28=(x + 7)(x - 4) = 0?
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Hi Schachfreizeit
Thanks for your query.


After looking at your post, I understand that you are unaware of the concept of solving quadratic equations using the “splitting the X-coefficient" method. Through this response, I’ll first introduce you to this concept, and then you’ll understand exactly what Brent did. 😊


INTRODUCTION TO CONCEPT (Splitting the X-Coefficient):
Consider the equation: \(x^2 + 5x + 6 = 0\).

Now, how does this method actually work? Let us see how!

Our quadratic equation here is \(x^2 + 5x + 6 = 0\). Now, let us go step-by-step to learn this approach.

Step 1: Express the constant term as a product of two numbers.
Note that there can be multiple ways of doing this. For example, here, the constant term = 6 [\(x^2 + 5x + 6 = 0\)]
  • So, the possible ways of expressing 6 as a product of two numbers are:
    • (1 × 6); (2 × 3); (-1 × –6) and (-2 × –3).

Step 2: From the possible pairs of numbers obtained in Step 1, choose the pair whose sum equals the coefficient of x.
Here, coefficient of x = +5 [\(x^2\) + 5x + 6 = 0]
  • Pair 1 is (1, 6):
    • 1 + 6 = 7 ≠ 5
  • Pair 2 is (2, 3):
    • 2 + 3 = 5 = 5 (BINGO! No need to check further because no other pair will work. Check for yourself!)
Only 2 and 3 are possible.

Step 3: Split the coefficient of x into a sum of two numbers – the exact numbers found in Step 2.
That is, \(x^2 + 5x + 6 = x^2 + 2x + 3x + 6\)

Step 4: Take out common factors from the 4 terms obtained in Step 3 – separately from the first two terms and the last two terms. Continue taking common as shown below.
  • x^2 + 2x + 3x + 6
  • = x (x + 2) + 3 (x + 2) --- [Pairwise common]
  • = (x + 2) (x + 3) ---- [(x + 2) common from both new terms]


And that’s it! Now, you can write \(x^2 + 5x + 6 = (x + 2) (x + 3) = 0\).
So, after seeing this explanation, I would suggest you to take a while here and try by yourself to convert x² + 3x – 28 to (x + 7) (x – 4).

I hope you have tried converting the above equation. Let me write the steps for you to verify your calculations.


SOLUTION TO ORIGINAL QUESTION:
We have to solve \(x^2 + 3x – 28 = 0\).
Solution:
Step 1: Splitting – 28 into a product of two numbers:
    (- 1 × 28); (1 × - 28); (- 2 × 14); (2 × - 14); (- 4 × 7); (4 × - 7).
Step 2: Choosing the pair whose sum = x-coefficient = 3.
    Only – 4 + 7 gives 3. So, chosen pair is (- 4 × 7).
Step 3: Rewriting the equation \(x^2 + 3x – 28\) as \(x^2 – 4x + 7x – 28\)
Step 4: Taking pair-wise common and then overall common:
    \(x^2 – 4x + 7x – 28 = x (x – 4) + 7(x – 4) = (x + 7) (x – 4)\)


And that’s it! This is the entire process. And if you proceed one step further from here, you can easily say that the roots of this equation are –7 and 4.


PRACTICE QUESTIONS:
Here are a few equations for you to practice. 😊
1. Solve \(x^2 + 10x + 16 = 0\).
2. Solve \(x^2 – 6x + 8 = 0\).
3. Solve \(x^2 + x – 12 = 0\).


Hope this helps!


Best,
Aditi Gupta
Quant expert, e-GMAT
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