We can write the given equation 8x^{2} - 22x - 21 = 0
as 8x^{2} – 28 x + 6x - 21 = 0, which is the method of splitting the middle term.
Or 4x (2x – 7) + 3 (2x – 7) = 0,
(4x + 3) * (2x – 7) = 0,
So we have converted the Quadratic Equation in the form of the factors. So we get
Either 4x + 3 = 0 or 2x – 7 = 0,
Or 4x = -3 or 2x = 7,
X = -3 /4 or x = 7 / 2.
These are the two roots of the given quadratic equation.
The given equation is in the form of 2x^{2} + kx + 3 = 0, so we say that this equation is in the form of Quadratic Equation ax^{2} + bx + c = 0.
So here we have a = 2, b = k and c = 3, so putting the values in D= b^{2} – 4 * a * c, we get:
D = b^{2} – 4 * a *c,
So D= k^{2} – 4 * 2 * 3,
Or D = k^{2} – 24.
Further we know that if the roots are real and equal, then value of 'D' will be zero, so we write:
k^{2} – 24 = 0,
or k^{2} = 24,
or k = + √24 or k = - √24,
or k = +2 √6 or k = - 2 √6 are two values of 'k' for which we find that given equation will have real and equal roots.
The given equation is in the form 4x^{2} – 3 * k *x + 1 = 0, so we say that this equation is in the form of the Quadratic Equation ax^{2} + bx + c = 0.
So here we have a = 4, b = -3k and c = 1, so putting the values in D= b^{2} – 4 * a * c, we get:
D = (9k^{2} – 4 * 4 * 1) = 9 k^{2} – 16.
Further we know that if roots are real and equal, then value of 'D' will be zero, so we write:
9 k^{2} – 16 = 0,
Or 9k^{2} = 16,
Or k^{2} = 16 / 9.
So taking Square root on both sides of the equation we get
k = 4/3 or k = - 4/3, there are two values of 'k' for which we find that the given equation will have real and equal roots.
We have the given equation 2 x^{2} – 6x + 3 = 0 and we observe that the equation is of the form ax^{2} + bx + c = 0. Here we observe that on comparing the given equation with standard equation, we get : a = 2, b = -6, c = 3.
Now we will find the value of D = b^{2} – 4 * a * c,
= (-6)^{2} – (4 * 2 * 3),
= 36 – 24,
= 12.
Now if we look at the value of 'D', it is greater than zero, so we say that it is a positive number. So we come to the conclusion that the roots of the equation are real and unequal. Now solving the equation by formula of the Quadratic Equation, we will apply the formula for finding the roots of equation as follows :
x = -b + root ( D) / 2 *a or X = -b - root ( D) / 2 *a,
x = (-(-6) +√12) / 2 * 2 or x = (-(-6) - √12) / 2 * 2,
Or x =( 6 +√12) / 4 or Or x =(6 - √12) / 4,
Or we write x = (6 +2√3) / 4 or Or x =(6 - 2 √3) / 4,
Or we write x = (3 +√3) / 2 or Or x =( 3 - √3) / 2.
We observe that above equation is in the form of ax^{2} + bx + c = 0, so given equation is 6x^{2} - x – 2 = 0 can be solved as follows : 6x^{2} - 4x + 3x - 2 = 0.
Now we will take 2x common from first 2 terms and we get :
2x * (3x – 2) + 1 * (3x – 2) = 0,
Now we will take (3x – 2) common from two Sets of terms and we get:
(2x + 1) * (3x – 2) = 0,
So we get:
(2x + 1) = 0 or (3x – 2) = 0,
2x = -1 or 3x = 2,
x = -1 /2 or x = 2 /3 are roots of given equation.
These roots can also be calculated by first finding the value of the value of 'D', which will help us to check if the value of the roots are real, imaginary , equal or unequal. Once we use the formula of D = b^{2} – 4 * a * c, we will be able to find the roots by x = -b + root ( D) / 2 *a or x = -b - root ( D) / 2 *a.
Given equation is:
5A =
Now we have to solve this equation for the variable ‘c’. First cross multiply the given equation. i.e.
25A =
25A = b + c;
Now subtract b on both side of the equation.
On subtracting we get.
25A – b = b + c – b;
25A – b = c;
We get the value of c = 25A– b.
A = (b + c) /3;
A = (b + c)
3
Now we have to solve this equation for the variable ‘c’. First cross multiply the given equation. i.e.
3A = (3) ( b + c)
(3)
3A = b + c;
Now subtract ‘b’ from both sides of the equation.
On subtracting we get.
3A – b = b + c – b;
3A – b = c,
We get the value of c = 3A– b.
Given equation is am + b = c ; and we have to solve this literal equation
In this equation first we have to move the variable ‘b’ and the sign becomes the opposite.
am + b = c;
On moving the variable ‘b’ we get:
am = c – b;
Now we have to move the variable ‘a’ and get the new equation and opposite of multiply is divide.
On moving the variable ‘a’ we get;
am = c - b;
X =
Now we have to solve this equation for the variable ‘h’. firstly cross multiply the given equation. i.e.
4(X) =
,
4X = xh - yh;
Now take common in the like variable i.e.
4X = h(x – y);
Now divide the (x – y) on both side of the equation.
On dividing we get:
Given equation is -5m + n = 20; and 5m + 8n = 17,
In these equations when we add both the equations then one variable can be eliminated and we get the value of another variable. For finding the value of eliminated variable we have to put the value of another variable in one equation. In this method the coefficient of one variable is opposite of another.
For solving these equations we have to follow some steps: i.e.
-5m + n = 20; and 5m + 8n = 17;
In this equation the coefficient of ‘q’ is opposite so we will solve this equation by addition rule.
-5m + n = 20 ….(1)
5m + 8n = 17.…(2)
9n = 37;
Step 2: - Now find the variable ‘n’,
By solving we get the n = 37/9.
Step 3:- Using the value of ‘n’ we can find the value of ‘m’, then putting the value of ‘n’ in equation (1)
-5m + n = 20; put value of n = 37/9;
-5m + 37/9 = 20;
-5m = 20 – 39/9;
- 5m = 180 –39;
9
-5m = 219;
9
-m = 219;
45
m = -219/45.
So after solving this equation we get the value of m = -73/15 and value of n = 37/9.
We have 9p – 2q = 11; and 5p + 2q = 3 equation and we have to solve this.
In these equations when we add both the equations then one variable can be eliminated and we get the value of another variable. For finding the value of eliminated variable, we have to put the value of another variable in one equation. In this method the coefficient of one variable is opposite of another.
For solving these equations we have to follow some steps: i.e.
9p – 2q = 11; and 5p + 2q = 3;
In this equation the coefficient of q is opposite so solve this equation by addition rule.
9p – 2q = 11….(1)
+ 5p + 2q = 3….. (2)
14p = 14;
Step 2: -
By solving we get the p = 1.
Step 3:- Using the value of ‘p’ we will find the value of ‘q’, putting the value of ‘p’ in equation (1).
9p – 2q = 11; put value of p = 1;
9(1) – 2q = 11;
9 – 2q = 11;
- 2q = 11 – 9;
-2q = 19,
q = 19/2,
So after solving this equation we get the value of p = 1 and value of q = 19/2.
We observe that as the number of days reduces, then the more number of men will dig the same pond. So the problem is the situation of inverse proportion. Let the number of men is ‘x’,
So we can write 12 * 8 = 6 * x,
Or we write: x = 12 * 8 / 6,
Or x = 16 men. Thus we say that 16 men will finish the same work in 6 days.
Here we observe that the problem is of inverse proportion. So let us assume that the 7 pipes will fill the tank in ‘x’ minutes.
So we can write: 6 * 84 = 7 * x,
Or, x = 84 * 6 / 7,
Or, x = 72 minutes. Thus we observe that the time taken by the pipe to fill the tank is 72 minutes.
We observe that as the Numbers of cows have reduced, the number of days the field will be grazed will increase, so it is again inverse proportion. We write that let the number of days = x,
So, 35 * 8 = 20 * x,
Or, 35 *8/20 = x,
Or, 35 *2/5 = x,
Or, 7 * 2 = x,
Or ,14 = x.
We conclude from the above problem that as the number of men will increase, the time consumed will reduce. So we say that it is the problem of inverse variance. Let us assume that ‘x’ men work for 35 hrs.
So, 14 * 45 = x * 35,
Or 35 x = 14 * 45,
Or x = 14 * 45/35 = 18 men.