MathsGee is Zero-Rated (You do not need data to access) on: Telkom |Dimension Data | Rain | MWEB

0 like 0 dislike
22 views
The quadratic equations $x^{2}-6 x+a=0$ and $x^{2}-c x+6=0$ have one root in common. The other roots of the first and second equations are integers in the ratio $4: 3 .$ Then the common root is

1) 1
2) 4
3) 3
4) 2
| 22 views

0 like 0 dislike

(4)

Explanation

Let $\alpha$ and $4 \beta$ be roots of $x^{2}-6 x+a=0$ and

$\alpha, 3 \beta$ be the roots of $x^{2}-c x+6=0$, then $\alpha+4 \beta=6$ and $4 \alpha \beta=a$

$\alpha+3 \beta=c$ and $3 \alpha \beta=6$

we get $\alpha \beta=2 \Rightarrow a=8$

So the first equation is $x^{2}-6 x+8=0 \Rightarrow x=2,4$ If $\alpha=2$ and $4 \beta=4$ then $3 \beta=3$

If $\alpha=4$ and $4 \beta=2$, then $3 \beta=3 / 2$ (non-integer)

$\therefore$ common root is $x=2$

by Bronze Status (9,536 points)

0 like 0 dislike