I need some help, Im trying to solve the below but to no avail, would appreciate your guidance :)
Find the value of $k$ if the roots of $$3x^2+5x-k=0,$$ differ by two.
$\endgroup$ 14 Answers
$\begingroup$Outline:
Let the roots be $u,v$.
Use Vieta's formulas to find $u+v$ and $uv$.
Then use the identity $(u-v)^2 = (u+v)^2 - 4uv$.
Once you have the value of $(u-v)^2$ in terms of $k$, set it to $4$ and solve for $k$.
$\endgroup$ $\begingroup$Note that the polynomial factors as $(3x+a)(x+b)$. Then the roots are $a/3$ and $b$, which have to differ by two. If you multiply $(3x+a)(x+b)$ out and compare it to the original polynomial, you should have enough relationships to extract the answer.
$\endgroup$ $\begingroup$Let $r $ and $r+2$ be the roots.
Their sum is $$r+ (r+2)=\frac{-5}{3}$$
thus
$$r=\frac {-11}{6}. $$
Their product is
$$r (r+2)=\frac {-k}{3}.$$
thus $$k=\frac{11}{12} $$
the roots are $-\frac {11}{6} $ and $\frac 1 6$
$\endgroup$ 0 $\begingroup$From the quadratic formula, we know that if the equation $ax^2+bx+c=0$ has two distinct real roots, then their difference is $2\sqrt{b^2-4ac}$. Plugging in the values from your equation, this means that $2\sqrt{25+12k}=2$. Solve for $k$.
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