Question

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The difference of the two solutions of x^2+19x+c=0 is 15. Find c.

Answer 1

Answer:

34

Step-by-step explanation:

For a quadratic equation ax² + bx + c = 0, the sum of the roots is -b/a, and the product of the roots is c/a.

If we say the roots are p and q, then:

p + q = -19/1

pq = c/1

The difference of the roots is 15:

p − q = 15

Add to the first equation:

2p = -4

p = -2

Which means the value of q is -17.

So the value of c is (-2)(-17) = 34.

Answer 2

The value of $c$ is 34.

It is given that,

• Difference of the two solutions of $x^2+19x+c=0$ is 15.

Explanation:

Let $m$ and $n$ are two solutions of the given equation.

$m-n=15$           ...(i)

The sum of solutions of $ax^2+bx+c=0$ is $-\dfrac{b}{a}$. So,

$m+n=-\dfrac{19}{1}$

$m+n=-19$          ...(ii)

Adding (i) and (ii), we get

$2m=-4$

$m=-2$

Substitute $m=-2$ in (ii).

$-2+n=-19$

$n=-19+2$

$n=-17$

The product of solutions of $ax^2+bx+c=0$ is $\dfrac{c}{a}$. So,

$mn=\dfrac{c}{1}$

$(-2)(-17)=c$

$34=c$

Thus, the value of $c$ is 34.

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