(3x-5)(x+2)leqx^2-এর সমাধান করুন

$$( 3 x - 5 ) ( x + 2 ) \leq x ^ { 2 }$$

Solve for x

$$x\in \begin{bmatrix}-\frac{5}{2},2\end{bmatrix}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x-5 by x+2 and combine like terms.

$$3x^{2}+x-10\leq x^{2}$$

Subtract x^{2} from both sides.

$$3x^{2}+x-10-x^{2}\leq 0$$

Combine 3x^{2} and -x^{2} to get 2x^{2}.

$$2x^{2}+x-10\leq 0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$2x^{2}+x-10=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 2 for a, 1 for b, and -10 for c in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 2\left(-10\right)}}{2\times 2}$$

Do the calculations.

$$x=\frac{-1±9}{4}$$

Solve the equation x=\frac{-1±9}{4} when ± is plus and when ± is minus.

$$x=2 x=-\frac{5}{2}$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-2\right)\left(x+\frac{5}{2}\right)\leq 0$$

For the product to be ≤0, one of the values x-2 and x+\frac{5}{2} has to be ≥0 and the other has to be ≤0. Consider the case when x-2\geq 0 and x+\frac{5}{2}\leq 0.

$$x-2\geq 0 x+\frac{5}{2}\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-2\leq 0 and x+\frac{5}{2}\geq 0.

$$x+\frac{5}{2}\geq 0 x-2\leq 0$$

The solution satisfying both inequalities is x\in \left[-\frac{5}{2},2\right].

$$x\in \begin{bmatrix}-\frac{5}{2},2\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-\frac{5}{2},2\end{bmatrix}$$