Silly Mistakes

Algebra Level 3

Jason and Thompson were solving the quadratic equation $x^{2} + bx + c = 0$ .

Jason wrote down the wrong value of $b,$ and found the roots to be 6 and 1. Thompson wrote down the wrong value of $c,$ and found the roots to be $-4$ and $-1.$

What are the actual roots of the equation?

-6

and

-1

-2

and

-3

-4

and

-6

6

and

1

3 solutions

Rohit Ner
Jun 1, 2015

Let $\alpha$ and $\beta$ be the roots of the equation.

Jason miscopied the value of $b$ however he copied the correct value of $c$ . By Vieta's Formula , the product of the roots is $c,$ so the product of roots of the incorrect equation would be same as that of the given equation: $\alpha \cdot \beta = 6\cdot 1 = 6.$

Thompson miscopied the value of $c$ however he copied the correct value of $b$ . By Vieta's Formula , the sum of the roots is $-b,$ so the sum of roots of the incorrect equation would be same as that of the given equation: $\alpha + \beta = (-4)+(-1)= -5.$

Solving $\alpha+\beta = -5$ and $\alpha\cdot\beta =6,$ we find the actual roots as $\color{#3D99F6}{\huge\boxed {-2,-3}}.$

Very short and elegant sol ...I solved all the equations

Naman Kapoor - 6 years ago

Farah Roslend
Jun 20, 2015

(X-6)(X-1)=X^2-7X+6 => c=6

(X+4)(X+1)=X^2+5X+4 => b=5

The actual equation is then: X^2+5X+6=(X+2)(X+3)

So the actual roots must be -2,-3.

Very short and clear solution!!

Prasit Sarapee - 5 years, 6 months ago

Rahul Narayan
Jun 8, 2015

Since sum of roots is equal to -b/a and product of roots is equal to c/a (standard notations); henceforth -b/a=-5 and c/a=6. Since a=1 , the value of b and c comes to be -5 and 6 respectively. On solving the obtained equation the roots comes to be -2 and -3. :)

Silly Mistakes

3 solutions

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