Long Questions

Solve by graphically x² - 3x = 10.

Here,
Let, x² - 3x - 10 = y --- (i)

Comparing equation (i) with ax² + bx + c = y, we get,
a = 1, b = -3, c = -10
a is positive. So, the curve is concave upward.

Now, vertex of parabola,
(x, y) = $(\frac{-b}{2a}, \frac{4ac\text{ - } b^2}{4a})$ = $(\frac{3}{2}, \frac{\text{-40 - 9}}{4})$
(x, y) = (1.5, -12.25)

Finding passing points of parabola,

`x`	-1	-2	0	1	1.5	2	3	5
`y`	-6	0	-10	-12	-12.25	-12	-10	0

Finally, graph of given quadratic equation is,
graph of x square - 3x - 10 = y

The parabola cuts the x-axis at (-2, 0) and (5, 0). So, x = -2 and x = 5.

Solve by graphically x² + 2x - 3 = 0.

Here,
Let, x² + 2x - 3 = y --- (i)

Comparing equation (i) with ax² + bx + c = y, we get,
a = 1, b = 2, c = -3
a is positive. So, the curve is concave upward.

Now, vertex of parabola,
(x, y) = $(\frac{-b}{2a}, \frac{4ac\text{ - } b^2}{4a})$ = $(\frac{-2}{2}, \frac{\text{-12 - 4}}{4})$
(x, y) = (-1, -4)

Finding passing points of parabola,

`x`	-4	-3	-2	-1	0	1	2
`y`	5	0	-3	-4	-3	0	5

Finally, graph of given quadratic equation is,
graph of x square + 2x - 3 = y The parabola cuts the x-axis at (-3, 0) and (1, 0). So, x = -3 and x = 1.

Solve graphically the quadratic equation x² - 2x - 3 = 0.

Here,
Let, x² - 2x - 3 = y --- (i)

Comparing equation (i) with ax² + bx + c = y, we get,
a = 1, b = -2, c = -3
a is positive. So, the curve is concave upward.

Now, vertex of parabola,
(x, y) = $(\frac{-b}{2a}, \frac{4ac\text{ - } b^2}{4a})$ = $(\frac{2}{2}, \frac{\text{-12 - 4}}{4})$
(x, y) = (1, -4)

Finding passing points of parabola,

`x`	1	0	-1	2	-2	3	4
`y`	-4	-3	0	-3	5	0	5

Finally, graph of given quadratic equation is,
graph of quadratic equation x square - 2x - 3 = 0

The parabola cuts the x-axis at (-1, 0) and (3, 0). So, x = -1 and x = 3.

Solve by graphically x² - 3x + 2 = 0.

Here,
Let, x² - 3x + 2 = y --- (i)

Comparing equation (i) with ax² + bx + c = y, we get,
a = 1, b = -3, c = 2
a is positive. So, the curve is concave upward.

Now, vertex of parabola,
(x, y) = $(\frac{-b}{2a}, \frac{4ac\text{ - } b^2}{4a})$ = $(\frac{3}{2}, \frac{\text{8 - 9}}{4})$
(x, y) = (1.5, -0.25)

Finding passing points of parabola,

`x`	1.5	0	1	-1	2	3	4
`y`	-0.25	2	0	6	0	2	6

Finally, graph of given quadratic equation is,
graph of x square - 3x + 2 = y

The parabola cuts the x-axis at (2, 0) and (3, 0). So, x = 2 and x = 3.

Solve by graphically: x² = 5x - 6.

Here,
Let, x² - 5x + 6 = y --- (i)

Comparing equation (i) with ax² + bx + c = y, we get, a = 1, b = -5, c = 6
a is positive. So, the curve is concave upward.

Now, vertex of parabola,
(x, y) = $(\frac{-b}{2a}, \frac{4ac\text{ - } b^2}{4a})$ = $(\frac{5}{2}, \frac{\text{4(6) - }(-5)^2}{4})$
(x, y) = $(\frac{5}{2}, \frac{-1}{4})$ = (2.5, -0.25)

Finding passing points of parabola,

`x`	2.5	0	1	2	3	4	5
`y`	-0.25	6	2	0	0	2	6

Finally, graph of given quadratic equation is,
graph of x square - 5x + 6 = y The parabola cuts the x-axis at (2, 0) and (3, 0). So, x = 2 and x = 3.

Solve by graphical method: x² + x - 2 = 0

Here,
Let, x² + x - 2 = y --- (i)

Comparing equation (i) with ax² + bx + c = y, we get,
a = 1, b = 1, c = -2
a is positive. So, the curve is concave upward.

Now, vertex of parabola,
(x, y) = $(\frac{-b}{2a}, \frac{4ac\text{ - } b^2}{4a})$ = $(\frac{-1}{2}, \frac{\text{-8 - 1}}{4})$
(x, y) = $(\frac{-1}{2}, \frac{-9}{4})$ = (-0.5, -2.25)

Finding passing points of parabola,

`x`	-0.5	1	-1	2	-2	-3
`y`	-2.25	0	-2	4	0	4

Finally, graph of given quadratic equation is, graph of x square + x - 2 = y The parabola cuts the x-axis at (1, 0) and (-2, 0). So, x = 1 and x = -2.

y = x²; y = 2 - x

Given equations,
y = x² --- (i)
y = 2 - x --- (ii)

From equation (i),

`x`	0	±1	±2	±3
`y`	0	1	4	9

This gives parabola with its vertex at origin.

Again, from equation (ii),

`x`	0	1	2	3	4
`y`	2	1	0	-1	-2

This gives straight line.

Finally, graph of given equations is, graph of y = x square and y = 2 - x

From the graph, parabola and straight meets each other at (1, 1) and (-2, 4). So, the solutions are (1, 1) and (-2, 4).

x² - 2x - 8 = 0

Here,
Let, x² - 2x - 8 = y --- (i)

Comparing equation (i) with ax² + bx + c = y, we get,
a = 1, b = -2, c = -8
a is positive. So, the curve is concave upward.

Now, vertex of parabola,
(x, y) = $(\frac{-b}{2a}, \frac{4ac\text{ - } b^2}{4a})$ = $(\frac{2}{2}, \frac{\text{4(-8) - 4}}{4})$
(x, y) = (1, -9)

Finding passing points of parabola,

`x`	1	0	-1	2	-2
`y`	-9	-8	-5	-8	0

Finally, graph of given quadratic equation is,
graph of x square - 2x - 8 = y

The parabola cuts the x-axis at (-2, 0) and (4, 0). So, x = -2 and x = 4.

x² + 7x + 12 = 0

Here,
Let, x² + 7x + 12 = y --- (i)

Comparing equation (i) with ax² + bx + c = y, we get,
a = 1, b = 7, c = 12
a is positive. So, the curve is concave upward.

Now, vertex of parabola,
(x, y) = $(\frac{-b}{2a}, \frac{4ac\text{ - } b^2}{4a})$ = $(\frac{-7}{2}, \frac{\text{4(12) - 49}}{4})$
(x, y) = $(\frac{-7}{2}, \frac{-1}{4})$ = (-3.5, -0.25)

Finding passing points of parabola,

`x`	-3.5	-1	-2	-3	-4	-5
`y`	-0.25	6	2	0	0	2

Finally, graph of given quadratic equation is,
graph of x square + 7x + 12 = y

The parabola cuts the x-axis at (-3, 0) and (-4, 0). So, x = -3 and x = -4.