Question

Cho phương trình . Dùng Vi-ét để tìm nghiệm x₂ rồi tìm giá trị của m

a, \(4x^2+3x-m^2+3m=0\) ( \(x\)₁ = -2)

b, \(3x^2-2(m-3)x+5=0\) ( \(x\)₁ = \(\dfrac{1}{3}\) )

Answer 1

Lời giải:

a) Theo định lý Vi-et:

\(\left\{\begin{matrix} x_1+x_2=\frac{-3}{4}\\ x_1x_2=\frac{-m^2+3m}{4}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} -2+x_2=\frac{-3}{4}\\ (-2)x_2=\frac{-m^2+3m}{4}\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x_2=\frac{5}{4}\\ (-2)x_2=\frac{-m^2+3m}{4}\end{matrix}\right.\)

\(\Rightarrow \frac{-m^2+3m}{4}=(-2).\frac{5}{4}=\frac{-10}{4}\)

\(\Rightarrow -m^2+3m=-10\)

\(\Leftrightarrow m^2-3m-10=0\Leftrightarrow (m-5)(m+2)=0\Rightarrow \left[\begin{matrix} m =5\\ m=-2\end{matrix}\right.\)

b)

Theo định lý Vi-et \(\left\{\begin{matrix} x_1+x_2=\frac{2(m-3)}{3}\\ x_1x_2=\frac{5}{3}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{1}{3}+x_2=\frac{2(m-3)}{3}\\ \frac{1}{3}x_2=\frac{5}{3}\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} \frac{1}{3}+x_2=\frac{2(m-3)}{3}\\ x_2=5\end{matrix}\right.\)

\(\Rightarrow \frac{2(m-3)}{3}=\frac{1}{3}+5=\frac{16}{3}\)

\(\Rightarrow 2(m-3)=16\Rightarrow m=11\)

Answer 2

Lời giải:

a) Theo định lý Vi-et:

\(\left\{\begin{matrix} x_1+x_2=\frac{-3}{4}\\ x_1x_2=\frac{-m^2+3m}{4}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} -2+x_2=\frac{-3}{4}\\ (-2)x_2=\frac{-m^2+3m}{4}\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x_2=\frac{5}{4}\\ (-2)x_2=\frac{-m^2+3m}{4}\end{matrix}\right.\)

\(\Rightarrow \frac{-m^2+3m}{4}=(-2).\frac{5}{4}=\frac{-10}{4}\)

\(\Rightarrow -m^2+3m=-10\)

\(\Leftrightarrow m^2-3m-10=0\Leftrightarrow (m-5)(m+2)=0\Rightarrow \left[\begin{matrix} m =5\\ m=-2\end{matrix}\right.\)

b)

Theo định lý Vi-et \(\left\{\begin{matrix} x_1+x_2=\frac{2(m-3)}{3}\\ x_1x_2=\frac{5}{3}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{1}{3}+x_2=\frac{2(m-3)}{3}\\ \frac{1}{3}x_2=\frac{5}{3}\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} \frac{1}{3}+x_2=\frac{2(m-3)}{3}\\ x_2=5\end{matrix}\right.\)

\(\Rightarrow \frac{2(m-3)}{3}=\frac{1}{3}+5=\frac{16}{3}\)

\(\Rightarrow 2(m-3)=16\Rightarrow m=11\)