quadratic equations-1

Quadratic EquationsIn one unknown(variable)

Algebra

SyllabusQuadratic equations in one unknown(variable)Solving by:

> Factorization>Formula

Nature of Roots>Two distinct roots, if b2-4ac >0 >Two equal roots, if b2-4ac =0 >No real root if b2-4ac <0

Solving problems

Quadratic equations

Table of ContentsIntroduction

>What is linear equation>What is quadratic equation>Definition of quadratic equation>Some points about quadratic equations

Solving quadratic equations by>Factorization>Formula

Discriminant, and nature of roots

Quadratic equations

IntroductionYou know what a linear equation in one unknown looks like. Its general form is ax+b =0 (a and b are known constants and x is unknown). It is called linear because the power of x is one. (x1 is generally written as x)And the solution for a linear equation ax+b =0 is given by x = -b/a.Before defining a quadratic equation, I will show you some examples of quadratic equations (where x is variable) :2x2+7x-31=03x2+11=02x2-3x=0

IntroductionQuadratic equations

If you observed the above equations, the highest power of x (the unknown/variable) is two (2) in each equation.2x2+7x-31=03x2+11=02x2-3x=0

Introduction

The power of x in this term is two (2)

The power of x in this term is one (1)The power of x in this term is zero (0) –

as -31 can be thought of as -31 .x0The power of x in this term is two (2)

There is no x term in this equationThe power of x in this term is zero (0) –

as 11 can be thought of as 11 .x0There is no constant term in this

equation

The power of x in this term is one (1)

The power of x in this term is two (2)

Hence the highest power of x (variable) in this equation is two (2)

Hence the highest power of x (variable) in this equation is two (2)Hence the highest power of x (variable)

in this equation is two (2)

All the above are quadrati c equati ons

Quadratic equations

Definition

An equation with one unknown (variable), in which the highest power of the variable is two, is called quadratic equation.Examples:3x2+4x+7=0 (There are 3 terms on LHS)4x2+5x=0 (There are 2 terms on LHS)2x2-50=0 (There are 2 terms on LHS)x2=4 (It is same as x2-4=0) (Solution: x=±2)

(There are 2 terms on LHS)x2=0 (only one term, solution : x =0, trivial)

IntroductionQuadratic equations

How to solve a quadratic equation?

Solving quadratic eqns.Quadratic equations

We all know how to solve a linear equation in one unknown(variable). But a quadratic equation is more complicated to solve. But if we can reduce a given quadratic equation into a set of linear equations, then we can also easily solve the quadratic equation. Now the problem is how to reduce .We have seen that the general form of quadratic equation is ax2+bx+c = 0.First let us try to solve some simple forms of quadratic equations:

Solving a simple quadratic equation-1

Quadratic equations

Quadratic equation without x term:Ex: x2=4⇒ x2-4 = 0 (All terms transferred to LHS. There is no x term)⇒ x2-(2)2 = 0 (This is in the form a2-b2)⇒ (x+2).(x-2) = 0 [ a∵ 2-b2 =(a+b).(a-b)]⇒ (x+2) =0 OR (x-2) = 0 (By zero product rule - if product of two factors is zero, then at least one factor is zero)

⇒ x = -2 OR x = 2Hence there are two solutions for this quadratic equation. In general for every quadratic equation there will be two solutions.

Yes! The equation is now reduced to a product of two linear expressions. Hence each can be easily solved separately. Click to Continue

Solving quadratic eqns.

Solving a simple quadratic equation-2

Quadratic equations

Quadratic equation without constant term:Ex: 4x2+5x=0⇒ x(4x+5) = 0 (Taking out the common factor x from eqn )⇒ x =0 OR (4x+5) = 0 (By zero product rule - if product of two factors is zero, then at least one factor is zero)

⇒ x = 0 OR x = -5/4Hence there are two solutions for this quadratic equation. In general for every quadratic equation there will be two solutions.

[Teaser: Do you think there will be, in general, 3 solutions to a cubic equation –Yes! You are right, there will be. And by extension there will be n solutions for a n-the degree equation]

Yes! The equation is now reduced to a product of two linear expressions. Hence each can be easily solved separately.

Click to continue


Points on quadratic equations

Quadratic equations

1) The standard form of a quadratic equation is ax2+bx+c = 0, where x is unknown (variable) and a, b, and c are all real number constants and a ≠ 0. [a, b, c are called coefficients of the quadratic equation]

2) A quadratic expression can be factorized into two linear expressions – as ax2+bx+c = (px+q)(rx+s), and hence any given quadratic equation can be changed to the form (px+q)(rx+s) =0. [How to factorize – we will see later]

3) Now Zero product rule is applied to solve the equation. For example if (x+3)(x-2) =0, then (x+3)=0 or (x-2)=0; which means x=-3 or x=2.

4) Every quadratic equation has two solutions, which are also called the roots of the equation.

If a =0, then the equation is reduced to bx+c=0, which is no longer quadratic equation, but simply a linear equation. Click to continue


Solving quadratic equations (General case)

Quadratic equations Solving quadratic eqns.

Now we will see how to solve a quadratic equation in the most standard form. We have already seen that the standard form of a quadratic equation is ax2+bx+c = 0. We try to solve this equation by factorizing it into two linear equations, and then solve each linear equation separately.There are two methods:By FactorizationBy FormulaFactorization method is suitable for simple equations, where the coefficients in the equation are simple and easy to factorize.Formulas method is suitable for all types of problems, and can be applied systematically to solve any equation, and directly gives solutions.

Solving quadratic equations by factorization

Quadratic equations Solving by Factorization

This is a trail and error method (that means it is not systematic).

It is suitable when the coefficients a, b and c are small & simple, and can be easily factorized by hand.

The factorization is done as follows: Factorize a.c into p and q such that b = p+q

Then, the quadratic expression can be easily factorized.

Examples will make the concept clear

Steps: Clear all fraction and brackets, if necessary Transpose all the terms to LHS to get the

equation into standard form Factorize the expression on the LHS Put each factor equal to zero and solve



Worked out examples-1Problem: 2x2-7x=39

Solution: 2x2-7x=39 (Given quadratic equation)

⇒ 2x2-7x-39=0 (All terms transferred to LHS)[Comparing with standard form we can see a=2; b=-7; c=-39)[ Now we try factorize a.c = 2x-39=-2x3x13 into p and q such that p+q=b :

p q p+q Remarks

2 -3x13=-39 2-39=-37 ≠ -7 (b)

-2 3x13=39 -2+39=37 ≠ -7 (b)

-2x3=-6 13 -6+13=7 ≠ -7 (b)

2x3=6 -13 6-13=-7 =-7 (b) [Right]


Worked out examples-1Problem: 2x2-7x=39

Solution: (continued)2x2-7x-39=0

⇒ 2x2 +6x-13x-39=0 (bx term is split into px+qx) ⇒ 2x(x +3)-13(x+3)=0

⇒ (x+3)(2x-13) [by taking out common factor (x+3)] ⇒ (x+3)=0 OR (2x-13) = 0 [By zero product rule] ⇒ x = -3 OR 2x =13

⇒ x = -3 OR x =13/2 [These are solutions of given quadratic equation][Tip:-It is always a good idea to check the correctness of solutions by substituting the answers back into original equation, and see if it is satisfied]

Taking out common factor 2x from the first two terms.Click to continue

Taking out common factor -13 from the last two terms

Click to continue

Yes! The equation is now reduced to a product of two linear expressions. Hence each can be easily solved separately.

Click to continue


Worked out examples-2Problem: Solve

Analysis: Before solving let us analyze the problem. The given problem is not a quadratic equation (at least in the given form). There are fractions here. Hence we try to eliminate fractions first. How do we do it? Take GCD for LHS fractions and make it a single fraction. On RHS convert the given mixed number 2½ into improper fraction of 5/2. Then cross multiply to eliminate fractions.


Worked out examples-2Problem: Solve

Solution:

By combining fractions on LHS – first take GCD of (x-1) and x for the resultant fraction.Click to continue

By cross multiplication.Click to continue


Worked out examples-2

Solution: (Continued) ⇒ 4x2-4x+2 = 5x2-5x (Simplifying both sides)⇒ -x2+x+2 =0 (By transferring all terms to LHS)⇒ x2-x-2 =0 (By multiplying both sides by -1, so we get

positive x2 term)⇒ x2-2x+ x -2 =0⇒ x(x-2) + (x-2)=0 (By taking x common from first two terms)⇒ (x+1)(x-2) = 0 (By taking x-2 common )⇒ (x+1) = 0 OR (x-2) =0 (By zero product rule)⇒ x=-1 OR x=2

Now this is in standard form of quadratic equation. By comparison the coefficients are a=1; b=-1; c=-2. Click to continue

Now factorize axc =(1)x(-2)=-2, into two factors p and q such that p+q = b (i.e. -1). Factors of -2 are :-1 and 2 (and -1+2 =1 ≠ -1 (wrong)1 and -2 (and 1-2 = -1 =-1 (b, hence OK)Click to continue



Problem: Find the quadratic equation whose solution set is (-2,3)

Analysis: It is a reverse(inverted) problem. Here we are given the pair of solutions (as a quadratic equation has two solutions), and we have to find the original quadratic equation.Here is a hint. See the solution procedure for a normal problem (i.e., quadratic equation is given, and you have to find the solution set). You can simply start at the bottom of that procedure and move up the steps.Consider the solution steps for the problem 2x2-7x-39=0, which you have already solved. I have given the steps, which are already done by you in the next slide. Now if the solution set is given the steps can be numbered from bottom to top as shown

Analysis of the Solution of previous problem: 2x2-7x-39=0 -Required equation

2x2 +6x-13x-39=0 -Step 62x(x +3)-13(x+3)=0 -Step 5 (Expanding)

(x+3)(2x-13)=0 -Step 4 (x+3)=0 OR (2x-13) = 0 -Step 3 x = -3 OR 2x =13 -Step 2 x = -3 OR x =13/2 -step 1[These are solutions of

given quadratic equation]Now you can see that for a inverted(reversed) problem, the solution steps can also be simply reversed. That is great, and this is much simpler than original (normal) problem. No great mathematics involved here. Have you noticed the reversed implies symbol



Click to continue



Back to the Problem: Find the quadratic equation whose solution set is (-2,3)

Solution:Solution set =(-2,3) [Given]

⇒ x=-2 OR x=3 ⇒ x+2=0 OR x-3=0 [Two linear equations] ⇒ (x+2).( x-3)=0 [Now we take the product of the two

linear equations, where as in the normal problem we factorize into two linear equations]

⇒ x2 +2x-3x-6=0 [Expanding the equation] ⇒ x2 -x -6=0 [Simplifying LHS] – required equation



Problem: Find the quadratic equation whose solution set is (α,β)

Analysis: It is same as previous problem, but we are trying to find a general solution Solution:Solution set = (α,β) [Given]

⇒ x= α OR x = β ⇒ x-α=0 OR x-β=0 [Two linear equations] ⇒ (x-α).( x-β)=0 [By taking the product of two linear equations] ⇒ x2 -αx- βx+αβ =0 [Expanding the equation] ⇒ x2 –(α+ β)x+αβ=0 [Simplifying LHS] – required equation

[It simply means that for given roots α and β, the quadratic equation is defined by the coefficients a=1, b= -(α+ β), and c = αβ]



Some examples of the above type problems

1) What is the quadratic equation whose solution set is (3,4) x2 –(3+4)x+3.4=0 [substituting α = 3, β=4]Ξ x2 –7x+12=0 [On simplification]2) What is the quadratic equation whose solution set is (-5,7) x2 –( - 5+7)x+(-5).(7)=0 [substituting α = -5, β=7]Ξ x2 –2x-35=0 [On simplification]Note:2) Be very careful while substituting α and β values and associated signs.3) In a descriptive problem do not write the answer by direct substitution of α

and β values, but show full derivation.


Worked out examples-5Problem: If x =2 and x=3 are roots of the equation 3x2 –2mx+2n=0; find the values of m and n.

Analysis: This is another slightly twisted problem. Here roots of the quadratic equation given, but coefficients are not given. Hence what we do, we substitute two root values in the given equation, giving us two equations. Now we have two equations in two unknowns (m and n ), and we already know how to solve them . Now we proceed to solve the problem



Problem: If x =2 and x=3 are roots of the equation 3x2 –2mx+2n=0; find the values of m and n.

Solution: Substituting the 1st root (i.e., 2) in the give equation we get3(2)2 – 2m(2) +2n=0

⇒ 12 – 4m + 2n =0 ⇒ 4m –2n =12 [By rearrangement]

Now substituting 2nd root (i.e., 3) in the given equation we get3(3)2 – 2m(3) +2n=0

⇒ 27 – 6m + 2n =0 ⇒ 6m –2n =27 [By rearrangement]

Now we have two equations (linear) in two unknowns - give us 2m=15 ⇒ m =15/2Substituting this value of m in eqn we get 4(15/2) -2n =12 ⇒ 2n = 18 ⇒ n=9; Hence m=15/2 and n=9

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Quadratic equations Solving by Formula

Solving quadratic equations by Formula

This is the general method that can be applied routinely for any quadratic equation.The roots of the quadratic equation ax2+bx+c = 0, where a ≠0 can be obtained by using the formula:


Solving quadratic equations by Formula

Proof: ax2+bx+c = 0 [Given]⇒4a(ax2+bx+c ) =4a(0) [Multiplying the entire eqn by 4a]

⇒ 4a2x2+4abx+4ac = 0 ⇒ (2ax)2+2(2ax)(b)+4ac =0

[4a2x2 is written as (2ax)2 and 4abx is written as 2(2ax)(b)] ⇒ (2ax)2+2(2ax)(b)+b2-b2+4ac =0

[b2 is added and subtracted to LHS] ⇒ (2ax+b)2 -b2+4ac =0 ⇒ (2ax+b)2 =b2-4ac

⇒ (2ax+b) = ±

⇒

This part of the expression is in the form of p2+2pq+q2,where 2ax =p and b = q, and hence this part can be written as (p+q)2

Click to continue



Problem: Solve the quadratic equation using formula: 5x2-2x-3= 0

Solution: Comparing the given equation with standard form, ax2+bx+c = 0 we get a = 5, b = -2, c = -3 Substituting these values in the formula

[Carefully substitute values in the formula, observing signs duly. Do not attempt short cuts when substituting values in the formula]

⇒x= 1 OR -3/5 It is wonderful that this

number within the square root is perfect square. But this

does not always happenHence, when b2-4ac >0, we have two

distinct real roots



Problem: Solve the quadratic equation using formula: x2+6x+9= 0

Solution: Comparing the given equation with standard form, ax2+bx+c = 0,we get a = 1, b = 6, c = 9,Substituting these values in the formula

⇒x= -3 Now the value b2-4ac is zero

Hence when the value b2-4ac is zero, there is only one root



Problem: Solve the quadratic equation using formula: 5x2-6x+7= 0

Solution: Comparing the given equation with standard form, ax2+bx+c = 0,we get a = 5, b = -6, c = 7 Substituting these values in the formula

⇒x has no real roots

Now the value b2-4ac is negative, and there is no square root of a negative number

Hence when the value b2-4ac is negative ,the quadratic equation has no solutions in the set of real numbers.



Problem: Solve the quadratic equation using formula: 2x2+6x+3= 0

Solution: Comparing the given equation with standard form, ax2+bx+c = 0,we get a = 2, b = 6, c = 3 Substituting these values in the formula

Now the value b2-4ac is positive, but it is not a perfect square

Hence when the value b2-4ac is positive, the quadratic equation has two real roots. But since b2-4ac is not a

perfect square, the roots are irrational

Quadratic equations Nature of roots

Nature of Roots

We have seen in the previous four examples that the nature of roots depends on the value of b2-4ac. Hence this expression b2-4ac is called Discriminant (D) of the given quadratic equation. We can summarize the nature of roots based on value of Discriminant (D) as follows:

You will see in the advanced lessons that when discriminant

D <0, the roots are called imaginary or complex

Tip:If you are asked to find the nature of roots of a given quadratic equation, don’t attempt to solve the equation to find roots. But simply compute b2-4ac to find the nature of roots.

b2-4ac =0 ⇒ The roots are real and equal (only one root)b2-4ac >0 ⇒ The roots are real and unequal (two roots)

1) b2-4ac is a perfect square r⇒ oots are also rational2) b2-4ac is not a perfect square r⇒ oots are also irrational

b2-4ac <0 ⇒ There is no real root Please note that both rational and irrational numbers are real numbers

Quadratic equations

You need not remember above results. You can simply and easily derive the above from the formula for equation of roots

Case I:B2-4ac >0

Case II:B2-4ac =0

Case III:B2-4ac <0

Then this expression evaluates to some positive value, say p. Then the roots are (-b+p)/2a and (-b-p)/2a – and hence two

real roots.

Sub case I(i) – b2-4ac is also a perfect square, then the highlighted expression also evaluates to an integer and hence the roots also become rational.Click to continue

Sub case I(ii) – b2-4ac is not a perfect square, then the highlighted expression evaluates to an irrational number and hence the roots also become irrational.Click to continue

Then the highlighted expression value, say p, becomes 0. Hence whether the roots (-b+0)/2 or (-b-0)/2 are same.

Hence only one root

Then the highlighted expression value, say p, can not be evaluated, as there are

no real suare roots for a negative number.

Nature of roots



Problem: Without solving the equation 7x2-9x+2= 0, comment on the nature of roots

Solution: Comparing the given equation with standard form, ax2+bx+c = 0,we get a = 7, b = -9, c = 2Substituting these values in the formula: Discriminant D = b2-4acD = (-9)2 – 4(7)(2)

⇒ D = 81 – 56 ⇒ D = 25

Since D >0; The roots are real and unequal (There are two distinct real roots)Also D is perfect square as 25 = 52, Hence roots are also rational.Hence the roots are rational, real, and unequal.





⇒ D = 169 – 96 ⇒ D = 73

Since D >0; The roots are real and unequal (There are two distinct real roots)Also D is not a perfect square as 73 can not be written as square of an integer, Hence roots are also irrational.Hence the roots are irrational, real, and unequal.





⇒ D = 100 – 100 ⇒ D = 0

Since D =0; The roots are real and equal (There is only one real root)Hence the roots are real and equal.



Problem: Without solving the equation 2x2+8x+9= 0, comment on the nature of roots

Solution: Comparing the given equation with standard form, ax2+bx+c = 0,we get a = 2, b = 8, c = 9Substituting these values in the formula: Discriminant D = b2-4acD = (8)2 – 4(2)(9)

⇒ D = 64 – 72 ⇒ D = -8

Since D < 0; There are no real rootsHence there are no real roots(The roots are imaginary).



Problem: Find the value of ‘m’ , if the roots of the following quadratic equation are equal: (4+m) x2 +(m+1) x + 1 =0

Analysis: In this type of problems, we are given a quadratic equation in which one/more coefficients (a, b, c) of the given equation are not given, but expressed in some unknown term. For example in the above equation coefficients a and b are expressed in terms of ‘m’, which is an unknown. But we are given a condition about the nature of roots. For example in this problem we are given that the roots of the equation are equal. Hence we are given that b2-4ac =0. Now by substituting the coefficients in this condition we get another equation (some times linear, sometimes quadratic]. Now we solve this new equation for the unknowns in the coefficients.



Problem: Find the value of ‘m’ , if the roots of the following quadratic equation are equal: (4+m) x2 +(m+1) x + 1 =0

Solution: Comparing the given equation with standard form, ax2+bx+c = 0,we get a = (4+m), b = (m+1), c = 1Roots of the quadratic equation are equal [Given]

⇒ b2-4ac =0 ⇒ (m+1)2 – 4(4+m)(1) = 0 [Substituting values of a,b,c in the eqn] ⇒ (m2+2m+1) – (16+4m) = 0 [Expanding terms]

⇒ m2 -2m – 15 =0 [By simplification][Now this is a quadratic equation in variable ‘m’. Hence solve it by usual methods – by factorization or by formula]By factorizationNow by Comparing the given equation with standard form, ax2+bx+c = 0, [contd…..]



Solution: [contd…..]we get a = 1, b = -2, c = -15Now Factorize (a.c) into two factors p and q such that p+q = ba.c = (1).(-15) = -15 (- 1x3x5) [displaying the prime factors]Possible values of p and q are

Hence given equation can be written as:m2 -5m + 3m– 15 =0 [Or you can also write as m2 + 3m -5m– 15 =0]

⇒ m(m-5)+3(m-5) = 0 [By taking m common from first two terms and 3 common from last two terms]

⇒ (m+3)(m-5) = 0 [By taking common the factor (m-5)] ⇒ m= -3 OR m=5 [ For both these values of ‘m’ the original

equation has equal roots]

p q p+q Remarks

-3 5 2 ≠ -2

3 -5 -2 = -2 Hence OKThis table is shown only for your clarity. In actual problem solving , you need not show these computations, but you can workout these in marginsClick to continue

It is a good idea to check the results, by actually substituting one/both values in the original equation , and see if indeed it has equal roots. Click to continue



Problem: Solve the following equation for x, and give your answer correct to 2 decimal digits:

Analysis:First the given problem has fractions. Hence first eliminate fractions. Combine fractions on LHS by taking GCD and then cross multiply both sides of eqn, and simplify the given eqn into a standard form.Next we are asked to solve the equation correct to 2 decimal places. That is we may get roots that are irrational (i.e., √2 or √5 etc). You may look up square root values from the tables.



Solution:

Comparing the given equation with standard form, ax2+bx+c = 0,we get a = 2, b = -7, c = -1Substituting these values in the formula We get

[contd….]



Solution: [contd….]

57 is not a perfect square, nor can any factors be taken out of the root sign (radical). Click to continue

Since these are approximate solutions, you can not check these values by substituting them into original equation. The result will not be exactly zero. Click to continue

Quadratic equations

Equations Reducible to quadratic equations [Advanced problems]

Some equations, though are not quadratic , can be reduced to quadratic equation by some suitable substitution of an expression with a new variable . Now we can solve the reduced equation . Then we have to solve the substituted expression replacing dummy variable. Examples will make it clear:Some clues for reducing a given equation into quadratic equation: All the powers of the given equation are either even or all are oddAn entire expression can be substituted by a dummy variable

Quadratic equations Equations Reducible to quadratic equations

Worked out examples-11 [Advanced Problem]

Problem: Solve 2x4-5x2+3 = 0

Analysis: The given equation is 4th degree equation, as the highest power of variable x is 4. But if you observe that the powers of x in various terms of the equation are – 4, 2, and 0. All are even powers. Hence we can reduce the above equation into a quadratic equation by introducing a dummy variable t = x2. Substitute t for x2. The equation then reduces to: 2t2-5t+3 =0 Now this is a quadratic equation in t. On solving this equation we get t = 1 or t = 3/2; [two solutions for dummy variable][Now we replace dummy variable with the original variable] Hence we get : x2=1 and x2=3/2 [two simple quadratic equations]Solving each we get x = ±1 and x=±√(3/2) [ No wonder we got 4 solutions in all, as the original equation is 4th degree one]



Problem: Solve 2x4-5x2+3 = 0

Solution:2x4-5x2+3 = 0 [Given]

⇒ 2(x2)2-5(x2)+3 = 0 [Writing equation in terms of x2] ⇒ 2t2-5t+3 = 0 [By substituting x2 = t] ⇒ 2t2-3t-2t+3 = 0 [Splitting -5t into -3t and -2t for factorization] ⇒ t(2t-3) - (2t-3) =0 [Factorizing] ⇒ (t-1)(2t-3) =0 [Factorizing]

⇒ t = 1 OR t = 3/2 [Solutions for dummy variable] ⇒ x2 = 1 OR x2 = 3/2 [Replacing dummy variable with original one] ⇒ x2 – 1=0 OR x2 -3/2 = 0 [Two quadratic equations] ⇒ (x +1)(x-1) = 0 OR [x+√(3/2)][x-√(3/2)]=0 [ ∵ a2-b2 = (a+b)(a-b)] ⇒ x = 1 OR x = -1 AND x = √(3/2) OR x = -√(3/2)


Worked out examples-12 [Advance problem]

Problem: Solve (x2+3x)2 -(x2+3x)-6 = 0

Analysis: If you expand the given equation, you get 4th degree equation. But without expanding ,you can observe that the expression (x2+3x) can be substituted by a dummy variable t, to reduce the given equation into quadratic form.Substitute t for (x2+3x). The equation then reduces to: t2-t-6. Now this is a quadratic equation. On solving this equation Given eqn can be factorized as (t-3)(t+2) =0we get t = 3 or t = -2; [two solutions for dummy variable]Now we solve (x2+3x)= 3 and (x2+3x)=-2 (two simple quadratic eqns)Solving each we get And x=-1 and -2 [The detailed steps of solution left to student as exercise]

Quadratic equations

Equations involving radicals (square roots) [Advanced problems]

Sometimes the given equation involves radicals (i.e., square roots etc). In such cases, the radicals must be eliminated by raising both sides of the equation to required power (exponent).For example if the given equation is:The above equation is rationalized (radicals eliminated) as follows:

Now this is quadratic equation, which can be solved by methods already studiedClick to continue

Always check the solutions of quadratic equations (obtained by rationalizing) in the original equation – as some solutions may not satisfy the original equation. They have to be discarded. Click to continue

Quadratic equations Equations involving radicals


Problem:

Analysis: Since the equation contains radicals (square root terms) we can rationalize the equation by taking square on both sides and simplifying it to a quadratic equation. But squaring always generates large terms. If you observe that the LHS has two terms, which are reciprocals of each other. That is if one term is y, then other term is 1/y. This substitution makes the equation simpler to solve. Hence solve this equation by substitution.



Solution:

[By taking GCD on LHS and converting mixed number into improper fraction on RHS]

⇒6(y2+1) = 13y [On cross multiplication] ⇒ 6y2- 13y+6 = 0 [ on rearranging – we get a quadratic equation] ⇒ 6y2- 9y-4y+6 = 0

[-13y is split for factorization as -9x-4 = 36, which is product of 6x6] ⇒ 3y(2y-3) – 2(2y-3) = 0 ⇒ (3y-2)(2y-3) =0 ⇒ y = 2/3 OR y = 3/2

[Now we can replace dummy variable for original variable][Contd……]



Solution: [Contd….]

⇒ 9x = 4-4x OR 4x = 9-9x [On cross multiplication] ⇒ 13x = 4 OR ⇒ 13x = 9 [On simplification] ⇒ x = 4/13 OR x = 9/13

Now check the results in the original equation Substitute x = 4/13

[Contd….]



Solution: [Contd….]

Similarly you can check that other solution also satisfies the original equation

quadratic equations-1

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Transcript of quadratic equations-1