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

Finally, graph of given quadratic equation is,

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

Finally, graph of given quadratic equation is,
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 y 1 0 -1 2 -2 3 4 -4 -3 0 -3 5 0 5

Finally, graph of given quadratic equation is,

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

Finally, graph of given quadratic equation is,

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

Finally, graph of given quadratic equation is,
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 y -0.5 1 -1 2 -2 -3 -2.25 0 -2 4 0 4

Finally, graph of given quadratic equation is, 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 y 0 ±1 ±2 ±3 0 1 4 9
This gives parabola with its vertex at origin.
Again, from equation (ii),
 x y 0 1 2 3 4 2 1 0 -1 -2
This gives straight line.

Finally, graph of given equations is,

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

Finally, graph of given quadratic equation is,

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

Finally, graph of given quadratic equation is,

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