Question

One of the roots of the equation 10x2−33x+c=0 is 5.3. Find the other root and the coefficient c.

Answer 1

Answer:

1. The other root is $-2$

2. c=-106

Step-by-step explanation:

The given equation is $10x^2-33x+c=0$.

If $5.3$ is a solution to the given quadratic equation, then $x=5.3$ must satisfy this equation.

$10(5.3)^2-33(5.3)+c=0$

$280.9-174.9+c=0$

$106+c=0$

$c=0-106=-106$

We now substitute $c=-106$ into the given equation to obtain;

$10x^2-33x-106=0$.

Comparing to the general equation;

$ax^2+bx+c=0$, we have a=10, b=-33 and c=-106.

Recall the quadratic formula;

$x=\frac{-b\pm \sqrt{b^2-4ac} }{2a}$

We now use the quadratic formula to obtain;

$x=\frac{--33\pm \sqrt{(-33)^2-4(10)(-106)} }{2(10)}$

This implies that;

$x=\frac{33\pm \sqrt{5329} }{20}$

$x=\frac{33\pm73}{20}$

$x=\frac{33-73}{20}$ or $x=\frac{33+73}{20}$

$x=\frac{-40}{20}$ or $x=\frac{106}{20}$

$x=-2$ or $x=5.3$

Hence the other root is $-2$