Quadratic Equation
Here `a` is the co-efficient of `x^2, b` is the co-efficient of `x` and the constant term is `c`.
Remember that: If `a=0` in `ax^2+bx+c=0`, then it reduces to a linear equation `bx+c=0`
The equation `x^2-7x+6=0` and `3x^2+4x=5` are examples of quadratic equations. `x^2-7x+6=0` is in standard form but `3x^2+4x=5` is not in standard form.
If `b=0` in a quadratic equation `ax^2+bx+c=0`, it is called a Pure Quadratic equation. For example `x^2-16=0` and `4x^2=7` are pure quadratic equations.
Solution of Quadratic Equations
To find the solution set of a quadratic equation, the following methods are used:
(i) Factorization (ii) Completing Square
(i) Solution by Factorization
In this method, write the quadratic equation in the standard form as `ax^2+bx+c=0`
If two numbers `r` and `s` can be found for an equation such that `r+s=b` and `rs = ac` then `ax^2+bx+c=0` can be factorized into two linear factors.
The procedure is explained in the following examples.
Example 01: Solve the quadratic equation `3x^2-6x=x+20` by factorization.
Solution: `3x^2-6x=x+20` -----(i)
The standard form of equation (i)
`3x^2-6x-x-20=0`
`3x^2-7x-20=0` -----(ii)
Here `a=3, b=-7, c=-20` and `ac=3\times-20=-60`
As `-12+5=-7 and -12\times5=-60`, so
The equation (ii) can be written as
`3x^2-12x+5x\-20=0`
`3x\left(x-4\right)+5\left(x-4\right)=0`
`\left(x-4\right)\left(3x+5\right)=0`
Either `x-4=0` or `3x+5=0`
`x=4` or `3x=-5`
or `x=-\frac(5)(3)`
`\therefore` `4` and `-\frac(5)(3)` are the roots of the given equation.
Thus, the solution set is `\{-\frac(5)(3),\4}`
Example 02: Solve `5x^2=30x` by factorization.
Remember that: Cancelling of `x` on both sides of `5x^2 = 30x` means the loss of one root i.e.,`x=0`
Solution: `5x^2=30x`
`5x^2-30x=0` which is factorized as
`5x(x-6)=0`
Either `5x=0` or `x-6=0`
`x=0` or `x=6`
`\therefore` `0` and `6` are the roots of the given equation.
Thus, the solution set is `\{0,6}`
(ii) Solution by Completing Square
To solve a quadratic equation by the method of completing the square is illustrated through the following examples.
Remember that: For our convenience, we make the co-efficient of `x^2` equal to 1 in the method of completing the square.
Example 01: Solve the equation `x^2-3x-4=0` by completing the square.
Solution: `x^2-3x-4=0` ----- (i)
shifting constant term `-4` to the right, we have
`x^2-3x=4`
Adding the square of `\frac(1)(2)\times` coefficient of `x`, that is,
`\left(-\frac(3)(2)\right)^2` on both sides of equation (ii), we get
`x^2-3x+\left(-\frac(3)(2)\right)^2=4+\left(-\frac(3)(2)\right)^2`
`\left(x-\frac(3)(2)\right)^2=4+\frac(9)(4)`
`\left(x-\frac(3)(2)\right)^2=\frac(16+9)(4)`
`\left(x-\frac(3)(2)\right)^2=\frac(25)(4)`
Taking the square root of both sides of the above equation, we get
`\sqrt{\left(x-\frac(3)(2)\right)^2}=\pm\sqrt{\frac(25)(4)}`
`x-\frac(3)(2)=\pm\frac(5)(2)`
`x=\frac(3)(2)\pm\frac(5)(2)`
Either `x=\frac(3)(2)+\frac(5)(2)` or `x=\frac(3)(2)-\frac(5)(2)`
`x=\frac(3+5)(2)` or `x=\frac(3-5)(2)`
`x=\frac(8)(2)` or `x=\frac(-2)(2)`
`x=4` or `x=-1`
Example 02: Solve the equation `2x^2-5x-3=0` by completing the square.
Solution: `2x^2-5x-3=0`
Dividing each term by 2,
to make the coefficient of `x^2` equal to 1.
`\frac(2x^2)(2)-\frac(5)(2)x-\frac(3)(2)`
`x^2-\frac(5)(2)x-\frac(3)(2)`
shifting constant term `\-frac(3)(2)` to the right
`x^2-\frac(5)(2)x=\frac(3)(2)` --------(i)
Multiply coefficent of `x` with `\frac(1)(2)` i.e., `\frac(1)(2)\left(-\frac(5)(2)\right)=-\frac(5)(4)`
Now adding `\left(-\frac(5)(4)\right)^2` on both sides of the equation (i), we have
`x^2-\frac(5)(2)x+\left(-\frac(5)(4)\right)^2=\frac(3)(2)+\left(-\frac(5)(4)\right)^2` , that is,
`\therefore` `a^2-2ab+b^2=\left(a-b\right)^2`
`\left(x-\frac(5)(4)\right)^2=\frac(3)(2)+\frac(25)(16)`
`\left(x-\frac(5)(4)\right)^2=\frac(24+25)(16)`
`\left(x-\frac(5)(4)\right)^2=\frac(49)(16)`
Taking square root of both sides of the above equation, we have
`\sqrt{\left(x-\frac(5)(4)\right)^2}=\pm\sqrt{\frac(49)(16)}`
`x-\frac(5)(4)=\pm\frac(7)(4)`
Either `x-\frac(5)(4)=\frac(7)(4)` or `x-\frac(5)(4)=-\frac(7)(4)`
`x=\frac(7)(4)+\frac(5)(4)` or `x=-\frac(7)(4)+\frac(5)(4)`
`x=\frac(7+5)(4)` or `x=\frac(-7+5)(4)`
`x=\frac(12)(4)` or `x=\frac(-2)(4)`
`x=3` or `x=-\frac(1)(2)`
`\therefore` `x=-\frac(1)(2), 3` are the roots of the given equation.
Thus the solution set is `{-\frac(1)(2), 3}`
Quadratic Formula
The quadratic equation in standard form is
ax^2+bx+c=0` `a\ne0`
Dividing each term of the equation by a, we get
`x^2+\frac (b)(a)x+\frac (c)(a)=0`
Shifting the constant term `\frac (c)(a)` to the right, we have
`x^2+\frac (b)(a)x=-\frac (c)(a)`
Adding `\left(\frac (b)(2a)\right)^2` on both sides, we obtain
`x^2+\frac (b)(a)x+\left(\frac (b)(2a)\right)^2=\left(\frac (b)(2a)\right)^2-\frac (c)(a)`
`x^2+\frac (b)(a)x+\left(\frac (b)(2a)\right)^2=\frac(b^2)(4a^2)-\frac (c)(a)`
or `\left(x+\frac (b)(2a)\right)^2=\frac(b^2-4ac)(4a^2)`
Taking square root of both sides, we get
`\sqrt{\left(x+\frac (b)(2a)\right)^2}=\pm\sqrt{\frac(b^2-4ac)(4a^2)}`
`x+\frac (b)(2a)=\pm\frac(\sqrt{b^2-4ac})(2a)`
`x=-\frac (b)(2a)\pm\frac(\sqrt{b^2-4ac})(2a)`
`\Rightarrow` `x=\frac(-b\pm\sqrt{b^2-4ac})(2a)`
Thus, `x=\frac(-b\pm\sqrt{b^2-4ac})(2a)` , `a\ne0` is known as "Quadratic Formula".
(ii) Use of Quadratic Formula:
A quadratic formula is a helpful tool for solving all those equations which can or can not be factorized. The method to solve the quadratic equation by using the quadratic formula is illustrated through the following examples.
Example 01: Solve the quadratic equation `2+9x=5x^2` by using the quadratic formula.
Solution: `2+9x=5x^2`
The given equation in standard form can be written as
`5x^2-9x-2=0`
Compared with the standard quadratic equation `ax^2+bx+c=0`, we observe that
`a=5 , b=-9 , c=-2`
Putting the values of `a`, `b` and `c` in quadratic formula
`x=\frac(-b\pm\sqrt{b^2-4ac})(2a)` , we have
`x=\frac(-\left(-9\right)\pm\sqrt{\left(-9\right)^2-4\left(5\right)\left(-2\right)})(2\left(5\right))`
`x=\frac(9\pm\sqrt{81+40})(10)`
`x=\frac(9\pm\sqrt{121})(10)`
`x=\frac(9\pm11)(10)`
Either `x=\frac(9+11)(10)` or `x=\frac(9-11)(10)`
`x=\frac(20)(10)` or `x=\frac(-2)(10)`
`x=2` or `x=-\frac(1)(5)`
`\therefore` `2,-\frac(1)(5)` are the roots of the given equation.
Thus, the solution set is `{-\frac(1)(5) , 2}`.
Example 02: Using the quadratic formula, solve the equation `\frac(2x+1)(x=2)-\frac(x-2)(x+4)=0`
Solution: `\frac(2x+1)(x=2)-\frac(x-2)(x+4)=0`
Simplifying and writing in the standard form
`\frac(2x+1)(x+2)=\frac(x-2)(x+4)`
By cross multiplication, we get
\`\left(2x+1\right)\left(x+4\right)=\left(x-2\right)\left(x+2\right)`
`\left(2x+1\right)\left(x+4\right)-\left(x-2\right)\left(x+2\right)=0`
`2x^2+8x+x+4-\left(x^2-2^2\right)=0`
`2x^2+9x+4-x^2+4=0`
`x^2+9x+8=0`
here `a=1` , `b=9` , `c=8`
By Using Quadratic Formula: `x=\frac(-b\pm\sqrt{b^2-4ac})(2a)`
`x=\frac(-9\pm\sqrt{\left(9\right)^2-4\left(1\right)\left(8\right)})(2\left(1\right))`
`x=\frac(-9\pm\sqrt{81-32})(2)`
`x=\frac(-9\pm\sqrt{49})(2)`
`x=\frac(-9\pm7)(2)`
Either `x=\frac(-9+7)(2)` or `x=\frac(-9-7)(2)`
`x=\frac(-2)(2)` or `x=\frac(-16)(2)`
`x=-1` or `x=-8`
`\therefore` `-1, -8` are the roots of the given equation.
Thus, the solution set is `{-8, -1}`.
Equations reducible to quadratic form
We now discuss various types of equations that can be reduced to a quadratic equation by some suitable substitution.
Type (i) The equations of the type `ax^4+bx^2+c=0`
Replacing `x^2=y` in equation `ax^4+bx^2+c=0`.
we get the quadratic equation in `y`.
Example 01: Solve the equation `x^4-13x^2+36=0`
Solution: `x^4-13x^2+36=0` -----------(I)
Let `x^2=y`, then `x^4=y^2`
Equation (i) becomes
`y^2-13y+36=0` which can be factorized as
`y^2-9y-4y+36=0`
`y\left(y-9\right)-4\left(y-9\right)=0`
`\left(y-9\right)\left(y-4\right)=0`
Either `y-9=0` or `y-4=0`
`y=9` or `y=4`
put `y=x^2`
`x^2=9` or `x^2=4`
Taking square root on both sides
`\sqrt{x^2}=\pm\sqrt9` or `\sqrt{x^2}=\pm\sqrt4`
`x=\pm3` or `x=\pm2`
`\therefore` The solution set is `{\pm2, \pm3}`
Type (ii) The equations of the type `ap\left(x\right)+\frac (b)(p\left(x\right))=c`
Example 02: Solve the equation `2\left(2x-1\right)+\frac(3)(2x-1)=5`.
Solution: `2\left(2x-1\right)+\frac(3)(2x-1)=5` ----------(i)
Let `2x-1=y`
Then the equation (i) becomes
`2y+\frac(3)(y)=5`
`\frac(2y^2+3)(y)=5`
`2y^2+3=5y`
`2y^2-5y+3=0`
Using quadratic formula: `x=\frac(-b\pm\sqrt{b^2-4ac})(2a)`
`y=\frac(-\left(-5\right)\pm\sqrt{\left(-5\right)^2-4\left(2\right)\left(3\right)})(2\left(2\right))`
`y=\frac(5\pm\sqrt{25-24})(4)`
`y=\frac(5\pm\sqrt{1})(4)`
`y=\frac(5\pm1)(4)`
`y=\frac(5+1)(4)` or `y=\frac(5-1)(4)`
`y=\frac(6)(4)` or `y=\frac(4)(4)`
`y=\frac(3)(2)` or `y=1`
When `y=\frac(3)(2)` When `y=1`
`\left(\therefore y=2x-1\right)` `\left(\therefore y=2x-1\right)`
`2x-1=\frac(3)(2)` `2x-1=1`
`2x=\frac(3)(2)+1` `2x=1+1`
`2x=\frac(3+2)(2)` `2x=2`
`2x=\frac(5)(2)` `x=\frac(2)(2)`
`x=\frac(5)(2\times2)` `x=1`
`x=\frac(5)(4)`
Thus, the solution set is `{1, \frac(5)(4)}`
Type (iii) Reciprocal equations of the type:
`a\left(x^2+\frac(1)(x^2)\right)+b\left(x+\frac(1)(x)\right)+c=0` or `ax^4+bx^3+cx^2+bx+a=0`
An equation is said to be a reciprocal equation if it remains unchanged when `x` is replaced by `\frac(1)(x)`.
Replacing `x` by `\frac(1)(x)` in `ax^4+bx^3+cx^2+bx+a=0` , we have
`a\left(\frac(1)(x)\right)^4-b\left(\frac(1)(x)\right)^3+c\left(\frac(1)(x)\right)^2-b\left(\frac(1)(x)\right)+a=0` which is simplified as
`a-bx+cx^2-bx^3+ax^4=0`. We get the same equation.
Thus `ax^4-bx^3+cx^2-bx+a=0` is a reciprocal equation.
The method for solving the reciprocal equations is illustrated through an example.
Example 03: Solve the equation `2x^4-5x^3-14x^2-5x+2=0`.
Solution: `2x^4-5x^3-14x^2-5x+2=0`
Dividing each term by `x^2`
`\frac(2x^4)(x^2)-\frac(5x^3)(x^2)-\frac(14x^2)(x^2)-\frac(5x)(x^2)+\frac(2)(x^2)=0`
`2x^2-5x-14-\frac(5)(x)+\frac(2)(x^2)=0`
Rearrange,
`2x^2+\frac(2)(x^2)-5x-\frac(5)(x)-14=0`
`2\left(x^2+\frac(1)(x^2)\right)-5\left(x+\frac(1)(x)\right)-14=0` ------------(i)
Let `x+\frac(1)(x)=y` , then `x^2+\frac(1)(x^2)=y^2-2`
So equation (i) becomes
`2\left(y^2-2\right)-5y-14=0`
`2y^2-4-5y-14=0`
`2y^2-5y-18=0`
`2y^2-9y+4y-18=0`
`y\left(2y-9\right)+2\left(y-9\right)=0`
`\left(2y-9\right)\left(y+2\right)=0`
Either `2y-9=0` or `y+2=0`
As `y=x+\frac(1)(x)`, so we have
`2\left(x+\frac(1)(x)\right)-9=0` or `x+\frac(1)(x)+2=0`
`2x+\frac(2)(x)-9=0` or `\frac(x^2+1+2x)(x)=0`
`\frac(2x^2+2-9x)(x)=0` or `x^2+2x+1=0`
`2x^2-9x+2=0`
here, `a=2`,`b=-9`, `c=2` or here, `a=1`,`b=2`, `c=1`
By quadratic formula: `x=\frac(-b\pm\sqrt{b^2-4ac})(2a)`
`x=\frac(-\left(-9\right)\pm\sqrt{\left(-9\right)^2-4\left(2\right)\left(2\right)})(2\left(2\right))` or `x=\frac(-2\pm\sqrt{\left(2\right)^2-4\left(1\right)\left(1\right)})(2\left(1\right))`
`x=\frac(9\pm\sqrt{81-16})(4)` or `x=\frac(-2\pm\sqrt{4-4})(2)`
`x=\frac(9\pm\sqrt{65})(4)` or `x=\frac(-2\pm0)(2)`
or `x=-1`
Thus, the solution set is `{-1, \frac(9-\sqrt{65})(4), \frac(9+\sqrt{65})(4)}`
Type (iv) Exponential Equation:
In exponential equations, the variable occurs in the exponent.
The method of solving such equations is illustrated through an example.
Example 04: Solve the equation `5^{1+x}+5^{1-x}=26`
Solution: `5^{1+x}+5^{1-x}=26`
`5^1.5^x+5^1.5^{-x}=26`
`5.5^x+\frac(5)(5^x)-26=0` ----------(i)
Let `5^x=y`. Then equation (i) becomes
`5y+\frac(5)(y)-26=`
`\frac(5y^2+5-26y)(y)=0`
`5y^2-26y+5=0`
`5y^2-25y-y+5=0`
`5y\left(y-5\right)-1\left(y-5\right)=0`
`\left(y-5\right)\left(5y-1\right)=0`
Either `y-5=0` or `5y-1=0`
`y=5` or `5y=1`
or `y=\frac(1)(5)`
Put `y=5^x`
`5^x=5^1` or `5^x=5^-1`
`x=1` or `x=-1`
Thus, the solution set is `{-1, 1}`
Type (v) The equations of the type:
`\left(x+a\right)\left(x+b\right)\left(x+c\right)\left(x+d\right)=k`, where `a+b=c+d`
Example 05: Solve the equation `\left(x-1\right)\left(x+2\right)\left(x+8\right)\left(x+5\right)=19`
Solution: `\left(x-1\right)\left(x+2\right)\left(x+8\right)\left(x+5\right)=19`
or `[\left(x-1\right)\left(x+8\right)\left(x+2\right)\left(x+5\right)-19=0]`
`\left(\because-1+8=2+5\right)`
`\left(x^2+8x-x-8\right)\left(x^2+5x+2x+10\right)-19=0`
`\left(x^2+7x-8\right)\left(x^2+7x+10\right)-19=0`
Let `x^2+7x=y`, then equation (i) becomes
`\left(y-8\right)\left(y+10\right)-19=0`
`y^2+10y-8y-80-19=0`
`y^2+2y-99=0`
`y^2+11y-9y-99=0`
`y\left(y+11\right)-9\left(y+11\right)=0`
`\left(y+11\right)\left(y-9\right)=0`
Either `y+11=0` or `y-9=0`
For: `y+11=0`
Put `y=x^2+7x`, so
`x^2+7x+11=0`
here, `a=1`, `b=7`, `c=11`
By using quadratic forumla, `x=\frac(-b\pm\sqrt{b^2-4ac})(2a)`
`x=\frac(-7\pm\sqrt{\left(7\right)^2-4\left(1\right)\left(11\right)})(2\left(1\right))`
`x=\frac(-7\pm\sqrt{49-44})(2)`
`x=\frac(-7\pm\sqrt{5})(2)`
For: `y-9=0`
Put `y=x^2+7x`, so
`x^2+7x-9=0`
or here, `a=1`, `b=7`, `c=-9`
By using quadratic forumla, `x=\frac(-b\pm\sqrt{b^2-4ac})(2a)`
`x=\frac(-7\pm\sqrt{\left(7\right)^2-4\left(1\right)\left(-9\right)})(2\left(1\right))`
`x=\frac(-7\pm\sqrt{49+36})(2)`
`x=\frac(-7\pm\sqrt{85})(2)`
Thus, the solution set is `{\frac(-7\pm\sqrt{5})(2), \frac(-7\pm\sqrt{85})(2)}`
Radical Equations
An equation involving expression under the radical sign is called a Radical Equation.
e.g., `\sqrt{x+3}=x+1` and `\sqrt{x-1}=\sqrt{x-2}+1`
(i) Equations of the type: `\sqrt{ax+b}=cx+d`
Example 01: `\sqrt{3x+7}=2x+3`
Solution: `\sqrt{3x+7}=2x+3` ------(i)
Squaring on both sides of equation (i)
`\left(\sqrt{3x+7}\right)^2=\left(2x+3\right)^2`
`3x+7=\left(2x\right)^2+2\left(2x\right)\left(3\right)+\left(3\right)^2`
`3x+7=4x^2+12x+9`
Simplifying the above equation:
`4x^2+9x+2=0`
Here, `a=4`, `b=9`, `c=2`
Applying quadratic formula
`x=\frac(-b\pm\sqrt{b^2-4ac})(2a)`
`x=\frac(-9\pm\sqrt{\left(9\right)^2-4\left(4\right)\left(2\right)})(2\left(4\right))`
`x=\frac(-9\pm\sqrt{81-32})(8)`
`x=\frac(-9\pm\sqrt{49})(8)`
`x=\frac(-9\pm7)(8)`
`x=\frac(-9+7)(8)` or `x=\frac(-9-7)(8)`
`x=\frac(-2)(8)` or `x=\frac(-16)(8)`
`x=\frac(-1)(4)` or `x=-2`
Checking:
Putting `x=\frac(-1)(4)` in the equation (i)
`\sqrt{3\left(-\frac(1)(4)\right)+7}=2\left(-\frac(1)(4)\right)+3`
`\sqrt{\frac(-3+28)(4)=-\frac(1)(2)+3`
`\sqrt{\frac(25)(4)=\frac(-1+6)(2)`
`\frac(5)(2)=\frac(5)(2)` which is true.
Putting `x=-2` in the equation (i)
`\sqrt{3\left(-2\right)+7}=2\left(-2\right)+3`
`\sqrt{-6+7}=-4+3`
`\sqrt1=-1`
`1=-1` which is not true.
On checking, we find that `x=-2` does not satisfy the equation (i), so it is an extraneous root.
Thus, the solution set is `{x=-\frac(1)(4)}`.
(ii) Equations of the type: `\sqrt{x+a}+\sqrt{x+b}=\sqrt{x+c}`
Example 02: Solve the equation`\sqrt{x+3}+\sqrt{x+6}=\sqrt{x+11}`
Solution: `\sqrt{x+3}+\sqrt{x+6}=\sqrt{x+11}` -------(i)
Squaring on both sides of the equation (i)
`\left(\sqrt{x+3}+\sqrt{x+6}\right)^2=\left(\sqrt{x+11}\right)^2`
`\left(\sqrt{x+3}\right)^2+2\left(\sqrt{x+3}\right)\left(\sqrt{x+6}\right)+\left(\sqrt{x+6}\right)^2=x+11`
`x+3+2\left(\sqrt{x+3}\right)\left(\sqrt{x+6}\right)+x+6=x+11`
`2x+9+2\left(\sqrt{x+3}\right)\left(\sqrt{x+6}\right)=x+11`
`2\left(\sqrt{x+3}\right)\left(\sqrt{x+6}\right)=x+11-2x-9`
`2\left(\sqrt{x+3}\right)\left(\sqrt{x+6}\right)==x+2` -------(ii)
Squaring on both sides of the equation (ii)
`[2\left(\sqrt{x+3}\right)\left(\sqrt{x+6}\right)]^2=\left(-x+2\right)^2`
`4\left(x+3\right)\left(x+6\right)=\left(-x\right)^2+2\left(-x\right)\left(2\right)+\left(2\right)^2`
`4\left(x^2+6x+3x+18\right)=x^2-4x-4`
`4\left(x^2+9x+18\right)=x^2-4x-4`
`4x^2+36x+72=x^2-4x-4`
`4x^2+36x+72-x^2+4x+4=0`
`3x^2+40x+68=0`
Here, `a=3`, `b=40`, `c=68`
Applying quadratic formula
`x=\frac(-b\pm\sqrt{b^2-4ac})(2a)`
`x=\frac(-40\pm\sqrt{\left(40\right)^2-4\left(3\right)\left(68\right)})(2\left(3\right))`
`x=\frac(-40\pm\sqrt{1600-816})(6)`
`x=\frac(-40\pm\sqrt{784})(6)`
`x=\frac(-40\pm28)(6)`
`x=\frac(-40+28)(6)` or `x=\frac(-40-28)(6)`
`x=\frac(-12)(6)` or `x=\frac(-68)(6)`
`x=-2` or `x=\frac(-34)(3)`
Checking:
Putting `x=\frac(-34)(3)` in the equation (i)
`\sqrt{\frac(-34)(3)+3}+\sqrt{\frac(-34)(3)+6}=\sqrt{\frac(-34)(3)+11}`
`\sqrt{\frac(-34+9)(3)}+\sqrt{\frac(-34+18)(3)}=\sqrt{\frac(-34+33)(3)}`
`\sqrt{\frac(-25)(3)}+\sqrt{\frac(-16)(3)}=\sqrt{\frac(-1)(3)}`
`\sqrt{\frac(25)(3)\times\left(-1\right)}+\sqrt{\frac(16)(3)\times\left(-1\right)}=\sqrt{\frac(1)(3)\times\left(-1\right)}`
`\frac(5)({\sqrt3})\iota+\frac(4)({\sqrt3})\iota=\frac(1)({\sqrt3})\iota` which is not true.
As `\frac(-34)(3)` is extraneous root, so the solution set is `{-2}`.
(iii) Equations of the type: `\sqrt{x^2+px+m}+\sqrt{x^2+px+n}=q`
Example 03: Solve the equation `\sqrt{x^2-3x+36}-\sqrt{x^2-3x+9}=3`
Solution: `\sqrt{x^2-3x+36}-\sqrt{x^2-3x+9}=3`
Let `x^2-3x=y`
Then `\sqrt{y+36}-\sqrt{y+9}=3`
Squaring on both sides
`\left(\sqrt{y+36}-\sqrt{y+9}\right)^2=\left(3\right)^2`
`\left(\sqrt{y+36}\right)^2-2\left(\sqrt{y+36}\right)\left(\sqrt{y+9}\right)+\left(\sqrt{y+9}\right)^2=9`
`y+36+y+9-2\left(\sqrt{y+36}\right)\left(\sqrt{y+9}\right)=9`
`2y+45-2\left(\sqrt{\left(y+36\right)\left(y+9\right)}\right)=9`
`-2\left(\sqrt{\left(y+36\right)\left(y+9\right)}\right)=9-2y-45`
`-2\left(\sqrt{\left(y+36\right)\left(y+9\right)}\right)=-2y-36`
`-2\left(\sqrt{\left(y+36\right)\left(y+9\right)}\right)=-2\left(y+18\right)`
`\left(\sqrt{\left(y+36\right)\left(y+9\right)}\right)=\left(y+18\right)`
Again, squaring on both sides
`\left(\sqrt{y^2+9y+36y+324}\right)^2=\left(y+18\right)^2\`
`y^2+45y+324=\left(y\right)^2+2\left(y\right)\left(18\right)+\left(18\right)^2`
`y^2+45y+324=y^2+36y+324`
`y^2+45y+324-y^2-36y-324=0`
`9y=0`
`y=\frac(0)(9)`
`y=0`
As `x^2-3x=y`
so, `x^2-3x=0`
`x\left(x-3\right)=0`
Either `x=0` or `x-3=0`
or `x=3`
`\therefore` `x=0,3` are the roots of the equation.
Thus, the solution set is `{0,3}`
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