Answer to Question #261683 in Algebra for jall

Question

Solve the equation 4x-3 = 2x² + 5x -6. Please indicate what you graphed and what you got the calculator to do.

Answer 1

The graph of the quadratic function $y=2x^2+5x-6$ is parabola.

$$y=2x^2+5x-6$$

$$=2\bigg(x^2+2(\dfrac{5}{4})x+(\dfrac{5}{4})^2\bigg)-2(\dfrac{5}{4})^2-6$$

$$=2\big(x+\dfrac{5}{4}\big)^2-\dfrac{73}{8}$$

$Vertex:\big(-\dfrac{5}{4}, -\dfrac{73}{8}\big)$

$y$ -intersection: $x=0, y=2(0)^2+5(0)-6=-6$

Point $(0, -6)$

$x$ -intersection(s): $y=0, 0=2\big(x+\dfrac{5}{4}\big)^2-\dfrac{73}{8}$

$$2\big(x+\dfrac{5}{4}\big)^2=\dfrac{73}{8}$$

$$\big(x+\dfrac{5}{4}\big)^2=\dfrac{73}{16}$$

$$x+\dfrac{5}{4}=\pm\sqrt{\dfrac{73}{16}}$$

$$x=-\dfrac{5}{4}\pm\dfrac{\sqrt{73}}{4}$$

$$x_1=-\dfrac{5}{4}-\dfrac{\sqrt{73}}{4}$$

$$x_2=-\dfrac{5}{4}+\dfrac{\sqrt{73}}{4}$$

Point $\big(\dfrac{-5-\sqrt{73}}{4}, 0\big),$ Point $\big(\dfrac{-5+\sqrt{73}}{4}, 0\big).$

$x=1: y=2(1)^2+5(1)-6=1$

Point $(1, 1).$

$x=-2: y=2(-2)^2+5(-2)-6=-8$

Point $(-1, -9).$

The graph of the linear function $y=4x-3$ is a straight line.

$x=0: y=4(0)-3=-3$ .

Point $(0, -3)$

$y=0: 0=4x-3=>x=\dfrac{3}{4}$

Point $\big(\dfrac{3}{4}, 0\big)$

Points of intersection $(1, 1), (-1.5, -9)$

Check

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

$$1=1, True$$

$$2(-1.5)^2+5(-1.5)-6=4(-1.5)-3$$

$$-9=-9, True$$

Solve the equation algebraically

$$2x^2+5x-6=4x-3$$

$$2x^2+5x-6-4x+3=0$$

$$2x^2+x-3=0$$

$$D=(1)^2-4(2)(-3)=25$$

$$x=\dfrac{-1\pm\sqrt{25}}{2(2)}=\dfrac{-1\pm5}{4}$$

$$x_1=\dfrac{-1-5}{4}=-1.5$$

$$y_1=4(-1.5)-3=-9$$

$$x_2=\dfrac{-1+5}{4}=1$$

$$y_2=4(1)-3=1$$

$$x_1=-1.5, x_2=1$$