If the quadratic formula is used to solve 2x 2- 3x- 1 = 0, what are the solutions?
In algebra, polynomials are algebraic expressions with exponents of the variables every bit whole numbers. When a polynomial is equated to nada, nosotros get a polynomial equation. If a quadratic polynomial is equated to zero, then we can call it a quadratic equation. A quadratic equation is an equation of second degree.
A quadratic equation makes a \( \cup \)-shaped curve (parabola) if nosotros represent it graphically. There are always 2 solutions to a quadratic equation with either existent roots or imaginary roots. Let us learn in detail the dissimilar methods of solving quadratic equations.
Definition of Quadratic Equations
An equation of 2d-degree polynomial in one variable, such as \(x\) unremarkably equated to cypher, is called a quadratic equation. The coefficient of \({x^2}\) must not exist zero in a quadratic equation.
\(p(ten)\) is a quadratic polynomial, then \(p(x) = 0\) called a quadratic equation.
For example, \(3{x^two} + 2x + two = 0, – {x^2} + 6x + ane = 0,7{x^2} – 6x + iv = 0\)etc., are quadratic equations.
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Standard Course of a Quadratic Equation
The standard grade of a quadratic equation is given past \(a{x^2} + bx + c = 0\) where\(a,b,c\) are real numbers, \(a \ne 0\) and \(a\) is the coefficient of \({10^2},b\) is the coefficient of \(x\) and \(c\) is a constant.
Roots of Quadratic Equations
The values of the variable, like \(ten\) that satisfy the equation in one variable are called the roots of the equation. The roots of the quadratic equation may be real or imaginary.
If a quadratic polynomial is equated to zero, information technology becomes a quadratic equation. The values of \(x\) satisfying the equation are known as the roots of the quadratic equation. Annotation that the zeroes of the quadratic polynomial \(a{x^2} + bx + c\) and the roots of the quadratic equation \(a{10^2} + bx + c = 0\) are the aforementioned.
Methods to Solve the Quadratic Equations
There are some methods to solve the quadratic equation. They are,
i. Factorization method
2. Completing square method
3. Formula method
4. Graphical Method
Solving Quadratic Equation by Factorization Method
If we can factorize \(a{ten^2} + bx + c,\,a \ne 0,\) into a production of two linear factors, and so the roots of the quadratic equation \(a{ten^2} + bx + c = 0\) tin can exist constitute by equating each factor to cypher.
Steps to Solve Quadratic Equation Using Factorization
If we tin can factorize \(a{10^2} + bx + c = 0,\,a \ne 0,\) into a product of two linear factors, then the roots of the quadratic equation \(a{x^2} + bx + c = 0\) tin can exist institute past equating each factor to naught.
Allow u.s.a. accept an example and hash out it.
Consider a quadratic equation \({x^2} + 5x + 6 = 0.\)
one. Compare the given quadratic equation with the standard class \(a{ten^two} + bx + c = 0\) and observe the coefficients of \({ten^2},x\) and the abiding to become the values for \(a,b,c.\)
So, comparison \({x^2} + 5x + six = 0\) with \(a{ten^2} + bx + c = 0\) we get, \(a = 1,b = v,c = 6.\)
ii. At present we will separate \(b\) as the sum of two numbers such that the product of these two numbers \( = a \times c = air-conditioning\)
We can factorize a quadratic equation when \(b\) can be split in \(v\) and \(w\) or \(b = v + due west\) and \( = a\, \times c = ac.\)
And then, \({x^ii} + 5x + 6 = 0 \Rightarrow {ten^2} + (ii + iii)x + 6 = 0 \Rightarrow {x^two} + 2x + 3x + 6 = 0.\)
3. Detect the common factor of the first two and last two terms separately, such every bit the new two terms have the same common gene.
So, \(x\) the common gene of the first two terms \({x^two} + 2x.\) Therefore, taking \(x\) outside, nosotros get \({x^ii} + 2x = 10(x + 2).\)
\(three\) is the mutual factor of the last two terms \(3x + 6.\) Therefore, taking \(3\) outside, we go \(3x+6=3(x+2).\)
So, \({ten^2} + 2x + 3x + 6 = 0\) can exist written equally \(x(x + 2) + 3(x + 2) = 0.\)
4. If we accept out the same common factor, so we get the production of two linear polynomials.
Hither, we tin can see that \((x + 2)\) is the common factor in the new 2 terms.
So, we tin write \(10(ten + two) + 3(x + 2) = 0\,as\,(10 + 2)(ten + iii) = 0\)
5. At terminal, we will utilise the zero product dominion to find the zeros. Null product belongings says that when \(p \times q = 0\) and so either \(p = 0\,or\,q = 0\)
Therefore, \((x + 2) = 0,or(x + three) = 0.\)
vi. Hence, the solutions or roots of the quadratic equation \({x^two} + 5x + 6 = 0\) are \(x = – 2,x = – 3.\)
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Solving Quadratic Equation past Completing the Foursquare Method
A quadratic equation can be solved by the method of completing the square.
Steps to Solve Quadratic Equation past Completing the Square Method
Consider the quadratic equation, \(a{10^ii} + bx + c = 0,a \ne 0\).
Permit u.s.a. divide the equation by \(a\).
\({x^2} + \frac{b}{a}x + \frac{c}{a} = 0\)
Multiply and separate \(two\) to \(10\) term.
\( \Rightarrow {x^ii} + 2 \times \frac{b}{{2a}} \times ten + \frac{c}{a} = 0\)
At present, to make the perfect foursquare, we need to add together and decrease \({\left( {\frac{b}{{2a}}} \right)^2}\) from 50.H.S.
\( \Rightarrow {x^2} + ii \times \frac{b}{{2a}} \times x + {\left( {\frac{b}{{2a}}} \right)^ii} – {\left( {\frac{b}{{2a}}} \right)^2} + \frac{c}{a} = 0\)
\( \Rightarrow {\left( {x + \frac{b}{{2a}}} \right)^2} – {\left( {\frac{b}{{2a}}} \right)^ii} + \frac{c}{a} = 0\)
Transferring \( – {\left( {\frac{b}{{2a}}} \right)^two} + \frac{c}{a}\) from \(L.H.S\) to \(R.H.S\) we accept,
\( \Rightarrow {\left( {x + \frac{b}{{2a}}} \correct)^ii} = – \frac{c}{a} + {\left( {\frac{b}{{2a}}} \correct)^2}\)
Taking square root on both sides nosotros have,
\( \Rightarrow x + \frac{b}{{2a}} = \pm \sqrt { – \frac{c}{a} + {{\left( {\frac{b}{{2a}}} \right)}^2}} \)
\( \Rightarrow 10 = \pm \sqrt { – \frac{c}{a} + {{\left( {\frac{b}{{2a}}} \right)}^2}} – \frac{b}{{2a}}\)
For example, consider the quadratic equation \(2{x^2} + 8x + 3 = 0\)
Let us divide the equation by \(a\, = \,ii\)
\(ii\left( {{10^two} + \frac{8}{2}x + \frac{three}{2}} \right) = 0\)
\( \Rightarrow {x^ii} + 4x + \frac{three}{ii} = 0\)
\( \Rightarrow {x^2} + ii \times 2x + \frac{3}{2} = 0\)
Now, to make the perfect square, we need to add and subtract \({(ii)^two}\) from \(L.H.S.\)
\( \Rightarrow {10^ii} + 2 \times 2 \times x + {(ii)^2} – {(2)^ii} + \frac{3}{2} = 0\)
\( \Rightarrow {(x + 2)^ii} – {(2)^2} + \frac{3}{2} = 0\)
Transferring \( – {(2)^2} + \frac{3}{2}\) from \(L.H.S\,to\,R.H.South\) we take,
\( \Rightarrow {(ten + 2)^2} = {(two)^2} – \frac{iii}{2}\)
Taking square root on both sides we have,
\( \Rightarrow ten + 2 = \pm \sqrt {four – \frac{3}{ii}} \)
\( \Rightarrow x = \pm \sqrt {\frac{5}{two}} – ii\)
Hence, the required solution of the quadratic equation \(2{ten^2} + 8x + 3 = 0\) is \(x = \pm \sqrt {\frac{5}{two}} – 2\)
Solving Quadratic Equation by Formula Method
The roots of the quadratic equation \(a{x^2} + bx + c = 0\) are given by the quadratic formula
\(x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\)
For example, consider the quadratic equation \(3{x^2} – 5x + ii = 0\)
From the given quadratic equation \(a = three,\,b = – 5,\,c = two\)
The quadratic Equation formula is given past
\(ten = \frac{{ – b \pm \sqrt {{b^two} – 4ac} }}{{2a}}\)
\(x = \frac{{ – ( – five) \pm \sqrt {{{( – 5)}^2} – 4 \times 3 \times 2} }}{{2a}} = \frac{{ + v \pm \sqrt {25 – 24} }}{vi}\)
\( = x = \frac{{ – ( – 5) \pm \sqrt {{{( – 5)}^2} – four \times three \times ii} }}{{2 \times 3}} = \frac{{5 \pm \sqrt {25 – 24} }}{6}\)
\( = \frac{{five \pm \sqrt 1 }}{6} = \frac{{five \pm ane}}{six} = \frac{{five + 1}}{6},\frac{{v – 1}}{6} = \frac{6}{six},\frac{4}{6}\)
\( \Rightarrow 10 = 1\) or \(x = \frac{ii}{3}\)
Hence, the roots of the given quadratic equation are \( \Rightarrow x = 1\) or \(ten = \frac{2}{three}\)
Graphical Solution of Quadratic Equation
The standard form of a quadratic equation is \(a{x^2} + bx + c = 0,\) where \(a,b\) and \(c\) are real and \(a\,\, \ne \,\,0\)
Let us have an example and hash out \({ten^2} – 5x + 6 = 0\)
Considering \(p(x) = {x^2} – 5x + vi\)
A quadratic polynomial e'er gives a parabolic graph. In a polynomial, the value of \(x\) which is responsible to make \(p(x) = 0\) is called the zero of the polynomial. When the polynomial equated with zero, it becomes an equation. If the zeros of the quadratic polynomial are known so they can exist considered as the solutions of the equation which is formed by equating the polynomial to zero.
When \(x = + two,p(2) = {(2)^ii} – 5(2) + vi = 4 – 10 + 6 = – half dozen + 6 = 0\)
\(x = + three,p(3) = {(3)^2} – 5(3) + 6 = 9 – 15 + 6 = – 6 + 6 = 0\)
Hence, the zeros of the quadratic polynomial are \( + iii,\, + 2\)
If we represent the polynomial graphically and then, the graph of the polynomial cuts the \(ten – \)axis at \( + iii\,{\rm{and}}\, + 2\)
Let us have another example and discuss \({ten^2} – 2x + ane = 0\).
Considering \(p(x) = {x^ii} – 2x + 1\)
A quadratic polynomial always gives a parabolic graph.
When \(x = + i,p(1) = {(i)^ii} – 2(1) + 1 = 1 – 2 + ane = – ane + 1 = 0\)
Hence, the zeros of the polynomial are \( + one\)
If we correspond the polynomial graphically then, the graph of the polynomial touches the \(x – \)axis at \( + 1\).
Therefore, the solution of the equation is \( + 1\).
Solved Examples – Methods of Solving Quadratic Equations
Q.ane. Notice the roots of the quadratic equation \(6{10^2} – x – 2 = 0\)
Ans: We take \(6{x^2} – x – 2 = 0\)
\( \Rightarrow 6{ten^ii} + 3x – 4x – 2 = 0\)
\( \Rightarrow 3x(2x + ane) – 2(2x + 1) = 0\)
The roots of \(6{x^2} – x – 2 = 0\) are the values of \(x\) for which \((3x – two)(2x + ane) = 0\)
Therefore, \(3x – 2 = 0\,\,or\,2x\,\, + \,\,1 = 0\)
\(x = \frac{two}{3},x = \frac{{ – 1}}{two}\)
Hence, the roots are \(\frac{two}{3}\& \frac{{ – 1}}{2}.\)
Q.ii. Find the roots of the quadratic equation by using the formula method \(2{x^2} – 8x – 24 = 0\)
Ans: From the given quadratic equation \(a = ii,b = – 8,c = – 24\)
Quadratic equation formula is given past \(10 = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\)
\(x = \frac{{ – ( – 8) \pm \sqrt {{{( – 8)}^two} – 4 \times 2 \times ( – 24)} }}{{two \times ii}} = \frac{{eight \pm \sqrt {64 + 192} }}{4}\)
\(x = \,\frac{{eight \pm \sqrt {256} }}{4}Due south = \frac{{viii \pm sixteen}}{4} = \frac{{8 + 16}}{4},\frac{{viii – 16}}{four} = \frac{{24}}{4},\frac{{ – 8}}{four}\)
\( \Rightarrow x = 6,x = – 2\)
Hence, the roots of the given quadratic equation are \(6\& – ii.\)
Q.3. Observe the roots of the quadratic equation \({x^two} + 10x + 21 = 0\) by completing the square method.
Ans: \({x^2} + 10x + 21 = 0\)
\( \Rightarrow {x^2} + 10x = – 21\) (Subtracted \(21\) from both sides of the equation)
\( \Rightarrow {x^2} + 10x + 25 = – 21 + 25\) (Added \({\left( {\frac{b}{2}} \right)^two} = {\left( {\frac{{10}}{2}} \right)^2} = 25\) on both the sides of the equation)
\( \Rightarrow {(x + v)^2} = 4\) (Completed the square by using the identity \({(a + b)^2} = {a^2} + 2ab + {b^ii}\))
And then, take the square root on both sides.
\(x + 5 = \pm 2\)
\(ten = – 3,ten = – vii\)
Hence, the roots of the given quadratic equation are \( – three\& – vii.\)
Q.iv. Find the roots of the quadratic equation\(2{ten^2} + 8x + 3 = 0\) past completing the square method.
Ans: \( \Rightarrow 2{x^2} + 8x = – 3\) [Subtracted \(3\) from both sides of the equation]
\( \Rightarrow {x^2} + 4x = \frac{{ – three}}{two}\) (Divided both sides of the equation by \(two\))
\( \Rightarrow {x^2} + 4x+4 = \frac{{ – three}}{two} + 4\) [Added \({\left( {\frac{b}{2}} \right)^2} = {\left( {\frac{4}{2}} \right)^2} = iv\) on both the sides of the equation]
\(\Rightarrow(x+2)^{2}=\frac{v}{2}\) [Completed the foursquare by using the identity \((a+b)^{2}=a^{two}+2 a b+b^{ii}\)]
So, take the square root on both the sides \( \Rightarrow 10 + 2 = \pm \sqrt {\frac{5}{ii}} \)
\( \Rightarrow 10 = -2 \pm \sqrt {\frac{5}{ii}} \) is the required solutions.
Q.five. Find the roots of the quadratic equation \({x^two} + 3x – 10 = 0\) by factorization method.
Ans: We accept \({x^2} + 3x – ten = 0\)
\( \Rightarrow {x^ii} + 5x – 2x – ten = 0\)
\( \Rightarrow x(x + 5) – 2(10 + 5) = 0\)
\( \Rightarrow (x – ii)(x + 5) = 0\)
And so, the roots of \({10^2} + 3x – 10 = 0\) are the values of \(x\) for which \((x – ii)(x + 5) = 0\)
Therefore, \(ten – 2 = 0\,\,or\,x + 5 = 0\)
\(ten = 2, – v\,\)
Hence, the roots are \(2\& – v.\)
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Summary
In this commodity, nosotros discussed quadratic equation in the variable \(ten\) which is an equation of the form \(a{x^2} + bx + c = 0,\) where \(a,b,c\) are real numbers, \(a\, \ne 0\) Besides, we discussed the methods of solving the quadratic equations such as factorizing method, completing the square method, formula method etc.
PRACTICE QUESTIONS ON QUADRATIC EQUATIONS
FAQs on Methods of Solving Quadratic Equations
Q.1. What are \(5\) methods of solving a quadratic equation?
Ans: We can solve the quadratic equations by using different methods given below:
ane. By factorizing method
2. By completing the square method
iii. By using the quadratic formula
4. By using the graphical method
v. By using the trial and fault method
Q.2. How can you lot identify a quadratic equation?
Ans: An equation is a quadratic equation in the variable \(x\) if information technology is of the form \(a{x^2} + bx + c = 0\) where \(a,b,c\) are existent numbers, \(a\, \ne \,0\).
Q.3. Which method can you use to solve all quadratic equations?
Ans: We can not use factorizing method and completing square method for every quadratic equation as in that location are some constraints. Nosotros can apply the formula method to solve all quadratic equations.
The roots of the quadratic equation \(a{x^2} + bx + c = 0\) are given by:
\(x = \frac{{ – b \pm \sqrt {{b^two} – 4ac} }}{{2a}}\)
This is the quadratic formula for finding the roots of a quadratic equation.
Q.4. How many types of quadratic equations are at that place?
Ans: There are three types of quadratic equations:
1. Standard form: \(a{x^2} + bx + c = 0,a \ne 0\)
2. Factored form: \((10 – a)(x – b) = 0\)
3. Vertex form: \(a{(x – h)^two} + k = 0\)
Each course of a quadratic equation has specific importance. Therefore, identifying the benefits of each different course can brand it easier to sympathize and solve dissimilar situations.
Q.5. Can y'all use the quadratic formula for any quadratic equation?
Ans: Yes, nosotros tin utilise the quadratic formula for whatsoever quadratic equation.
Q.6. What is the standard grade of the quadratic equation?
Ans: The form \(a{ten^2} + bx + c = 0,\,a \ne 0\) is called the standard form of a quadratic equation.
We hope you lot detect this detailed article on methods of solving quadratic equations helpful. If you take any doubts or queries regarding this topic, experience free to inquire us in the comment section.
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