Question

If the equation x2 −3x+1= p(x−3) has equal roots, find the possible value(s) of p.

Answer 1

Answer: p = 5, p = 1

Explanation

Given

$x^2−3x+1=p\left(x−3\right)$

Since the roots are equal, then the discriminant must be 0:

$b^2-4ac=0$

where a, b and c are the coefficients of the quadratic expression in the form ax² + bx + c = 0. Then, we must rearrange our equation in the previous form.

0. Multiplying ,p, times the expression inside the parenthesis:

$x^2−3x+1=p\cdot x−3p$

2. Setting the expression to 0:

$x^2−3x-px+1+3p=0$

2. Simplifying the expression by grouping similar terms:

$x^2-(3+p)x+(1+3p)=0$

Then, in our case: a = 1, b = –(3 + p), and c = (1 + 3p).

Now, we can replace the value in the discriminant and solve for p:

$(-(3+p))^2-4(1)(1+3p)=0$$(-3-p)^2-4(1+3p)=0$$(-3-p)^2-4+12p=0$$p^2-6p+5=0$

Finally, we can use the General Quadratic Expression to solve for p:

$p_{1,2}=\frac{-(-6)\pm\sqrt{(-6)^2-4(1)(5)}}{2(1)}$$p_{1,2}=\frac{6\pm\sqrt{36-20}}{2}$$p_{1,2}=\frac{6\pm\sqrt{16}}{2}$$p_{1,2}=\frac{6\pm4}{2}$

Now, calculating both results:

$p_1=\frac{6+4}{2}=\frac{10}{2}=5$$p_1=\frac{6-4}{2}=\frac{2}{2}=1$