Quadratic Equations
Class10Mathematics1Chapter2Maharashtra Board
Notes
Topics to be learn :

Quadratic Equations : Introduction
Linear and quadratic polynomial : When the degree of polynomial is 1 it is called a linear polynomial and if degree of a polynomial is 2 it is called a quadratic polynomial.
Quadratic equation : The equation obtained by taking the value of a quadratic polynomial zero, is called a quadratic equation.
Examples :
In the following given polynomials
5x + 9, x^{2} + 3x −5, 3x − 7, 3x^{2} − 5x, 5x^{2}
 5x + 9 and 3x – 7 are linear polynomial
 x^{2} + 3x −5, 3x^{2} − 5x, and 5x^{2 }are quadratic polynomial
 x^{2} + 3x −5 = 0, 3x^{2} − 5x = 0, and 5x^{2 }= 0 are quadratic equations formed by taking the value of the quadratic polynomials zero.
In above equations
 Coefficients of x^{2} are 1, 3 and 5 these coefficients are non zero.
 Coefficients of x are 3, −5 and 0
 Constants terms are −5, 0 and 0
 Here constant term of second and third polynomial is zero.
Standard form of quadratic equation :
The equation involving one variable and having the maximum index of the variable 2 is called a quadratic equation.
 The equation ax^{2} + bx + c = 0 is called the standard or general form of quadratic equation. Here, a, b and c are real numbers and a ≠
 In the equation ax^{2} + bx + c = 0, if b = 0, then the equation becomes ax^{2} + c = 0. This is also a quadratic equation.
 Similarly, if c = 0, then ax^{2} + bx = 0 is a quadratic equation.
 If b = 0 and c = 0, then ax^{2} = 0 is a quadratic equation.
In the quadratic equation ax^{2} + bx + c = 0, the constants a, b, c are called the quadratic coefficient, the linear coefficient and the constant term respectively.
Examples:
(1) Complete the following table
Quadratic equation  general form  a  b  c 
x^{2} − 4 = 0  x^{2} − 0x − 4 = 0  1  0  −4 
y^{2} = 2y − 7  y^{2} − 2y + 7 = 0  1  −2  7 
x^{2} + 2x = 0  x^{2} + 2x +0 = 0  1  2  0 
(2) Decide which of the following are quadratic equations ?
(1) 3x^{2} − 5x + 3 = 0 (2) 9y^{2} + 5= 0 (3) m^{3} − 5m^{2} + 4 = 0 (4) (l + 2) (l − 5) = 0
Solution :
(1) In the equation 3x^{2} − 5x + 3 = 0, x is the only variable and maximum index of the variable is 2, ∴ It is a quadratic equation.
(2) In the equation 9y^{2} + 5= 0, y is the only variable and maximum index of the variable is 2, ∴ It is a quadratic equation.
(3) In the equation m^{3} − 5m^{2} + 4 = 0, m is the only variable but maximum index of the variable is not 2. ∴ It is not a quadratic equation.
(4) Simplifying equation (l + 2) (l − 5) = 0
∴ l (l − 5) + 2 (l − 5) = 0
∴ l^{2} − 5l + 2l − 10 = 0
∴ l^{2} − 3l − 10 = 0, In this equation l is the only variable and maximum index of the variable is 2. ∴ It is a quadratic equation.
Roots of a quadratic equation :
The values of the variable which satisfy the given quadratic equation are called the solutions or the roots of the quadratic equation.
Examples:
(1) Let put x = −6 in the polynomial x^{2} + 5x − 6
x^{2} + 5x − 6 = (−6)^{2} + 5 × (−6) − 6
= 36 − 30 − 6 = 0
∴ x = −6 is a solution of the equation.
Hence −6 is one root of the equation x^{2} + 5x − 6 = 0
(2) 2x^{2} − 7x + 6 = 0 check whether (i) x = \(\frac{3}{2}\) , (ii) x = −2 are solutions of the equations.
Solution :
(i) Put x = \(\frac{3}{2}\) in the polynomial 2x^{2} − 7x + 6
2x^{2} − 7x + 6 = \(2(\frac{3}{2})^27(\frac{3}{2})+6\)
= \(2×\frac{9}{4}\frac{21}{2}+6\)
= \(\frac{9}{2}\frac{21}{2}+\frac{12}{2}\) = \(\frac{921+12}{2}=0\)
∴ \(\frac{3}{2}\) is a solution of the equation.
(ii) Let put x = −2 in 2x^{2} − 7x +6
2x^{2} − 7x + 6 = 2(−2)^{2} − 7(−2) + 6
= 2 × 4 + 14 + 6
= 8 + 14 +6 = 28 ≠ 0
∴ x = −2 is not a solution of the equation.
(3) If x = 5 is a root of equation kx^{2} − 14x − 5 = 0 then find the value of k by completing the following activity.
Solution :
One of the roots of equation kx^{2} − 14x − 5 = 0 is 5.
∴ Now Let substitute x = 5 in the equation.
∴ k(5)^{2} – 14(5) − 5 = 0
∴ 25k − 70 − 5 = 0
25k – 75 = 0
25k = 75
∴ k = \(\frac{75}{25}\) = 3
Remember :
 ax^{2} + bx + c = 0 is the general form of equation where a, b, c are real numbers and ’a ’ is non zero.
 The values of variable which satisfy the equation are called solutions or roots of the equation.
Let’s Recall : Find the factors of the following polynomials.
(i) x^{2} − 4 x −5, (ii) 2m^{2} − 5 m (iii) a^{2} − 25
Answer :
(i) x^{2} − 4 x −5,
= x^{2} − 5 x + 1x −5 = x(x – 5) + 1(x – 5) = (x + 1)(x – 5) 
(ii) 2m^{2} − 5 m
= m(2m – 5)

(iii) a^{2} − 25
= (a)^{2} – (5)^{2} = (a + 5)(a – 5) 
Methods of Solving Quadratic Equations are : Factorisation method, Completing square method and Formula method.
Solutions of a quadratic equation by factorisation :
Factorisation method :
 Write the given equation in the form ax^{2} + bx + c = 0
 Find the two linear factors of the LHS of the equation
 Equate each linear factor to zero.
 Solve each equation obtained in 3 and write the roots of the given equation.
Examples :
(1) Solve the quadratic equation x^{2} + 8x + 15 = 0 by factorisation method :
Step 1 : Split the middle term 8x as 3x and 5x. [Because 3x × 5x = 15x^{2 }= x^{2} × 15.]
x^{2} + 5x+ 3x + 15 = 0
Step 2 : Find the factors of LHS
x(x + 3) + 5(x + 3) = 0
(x + 5)(x + 3) = 0
Step 3 : If the product of two numbers is zero, then at least one of them must be zero.
x + 3 = 0 or x + 5 = 0
Step 4 : Solve each linear equation.
x = −3 or x = −5
Step 5 : Write the answer.
The roots of the equation are −3, −5
(2) Solve the following quadratic equations by factorisation.
(i) m^{2} − 14 m + 13 = 0, (ii) 3x^{2} − x − 10 = 0 , (iii) 3y^{2} = 15 y, (iv) x^{2} = 3,
(v) 6\(\sqrt{3}\)x^{2} + 7x = \(\sqrt{3}\)
(i) m^{2} − 14 m + 13 = 0
∴ m^{2} − 13 m − 1m + 13 = 0
∴ m(m − 13) − 1(m − 13) = 0
∴ (m − 13) (m − 1) = 0
∴ m − 13 = 0 or m − 1 = 0
∴ m = 13 or m = 1
∴ 13 and 1 are the roots of the given quadratic equation.
(ii) 3x^{2} − x − 10 = 0
∴ 3x^{2} − 6x + 5x − 10 = 0
∴ 3x(x − 2) + 5(x − 2) = 0
∴ (3x + 5) (x − 2) = 0
∴ (3x + 5) = 0 or (x − 2) = 0
∴ x = \(\frac{5}{3}\) or x = 2
 \(\frac{5}{3}\) and 2 are the roots of the given quadratic equation.
(iii) 3y^{2}= 15 y
∴ 3y^{2} − 15 y = 0
∴3y(y − 5) = 0
∴ 3y = 0 or (y − 5) = 0
∴ y = 0 or y = 5
∴ 0 and 5 are the roots of quadratic equation.
(iv) x^{2} = 3
∴ x^{2} − 3 = 0
∴ x^{2} − ( \(\sqrt{3}\) )^{2} = 0
∴ (x + \(\sqrt{3}\) ) (x − \(\sqrt{3}\) ) = 0
∴ ( x + \(\sqrt{3}\) ) = 0 or (x − \(\sqrt{3}\) ) = 0
∴ x = \(\sqrt{3}\) or x = \(\sqrt{3}\)
∴ \(\sqrt{3}\) and \(\sqrt{3}\) are the roots of given quadratic equation.
(v) 6 x^{2} + 7x = \(\sqrt{3}\)
∴ 6\(\sqrt{3}\)x^{2} + 7x − \(\sqrt{3}\) = 0
∴ 6\(\sqrt{3}\)x^{2} + 9x − 2x − \(\sqrt{3}\) = 0
∴3\(\sqrt{3}\)x(2x + \(\sqrt{3}\) ) −1 (2x + \(\sqrt{3}\) ) = 0
∴ (2x + \(\sqrt{3}\) ) (3\(\sqrt{3}\)x − 1) = 0
∴2x + \(\sqrt{3}\) = 0 or 3\(\sqrt{3}\)x − 1 = 0
∴ 2x = \(\sqrt{3}\) or 3\(\sqrt{3}\)x = 1
∴ x = \(\frac{\sqrt{3}}{2}\) or x = \(\frac{1}{3\sqrt{3}}\)
∴ \(\frac{\sqrt{3}}{2}\) and \(\frac{1}{3\sqrt{3}}\) are the roots of the given quadratic equation.
and are the roots of given quadratic equation.
Multiplying coefficient of x^{2} and constant.
6\(\sqrt{3}\) × \(\sqrt{3}\) = 18 (= 9 × 2) 9 = 3\(\sqrt{3}\) × \(\sqrt{3}\) 
Remember : While solving quadratic equation ax^{2} + bx + c , the coefficient of x^{2} i.e ‘a’ required to multiply by constant ‘c’ for factorisation
Solution of a quadratic equation by completing the square :
The quadratic equation of the type x^{2} + 10x + 2 = 0 cannot be solved by the method of factorisation, because we cannot ﬁnd two factors of 2 whose sum is 10.
In such a case, completing square method is used for solving quadratic equations.
For solving quadratic equation by this method proceed as follows :
Step 1 : Write the given equation in the form ax^{2} + bx + c = 0
x^{2} + 10x + 2 = 0
Step 2 : Considering the first two terms on LHS, find the third suitable square term to make the polynomial a perfect square.
Lets add a suitable term in first two LHS terms i.e. k
x^{2} + 10x + k = (x + a)^{2}
then x^{2} + 10x + k = x^{2} + 2ax + a^{2}
∴ 10 = 2a and k = a^{2}
∴ a = 5, ∴ k = a^{2} = (5)^{2} = 25
Step 3 : Add the square term and subtract the same.
x^{2} + 10x + 25 – 25 + 2 = 0
Step 4 : Write the square of the ﬁrst three terms and the last two terms.
(x + 5)^{2} − 25 + 2 = (x + 5)^{2} – 23 = (x + 5)^{2} – (\(\sqrt{23}\))^{2}
∴ (x + 5)^{2} – (\(\sqrt{3}\))^{2 }= 0
Step 5 : Factorise and equate each factor to zero.
∴ (x + 5 + \(\sqrt{3}\) )(x + 5 − \(\sqrt{3}\) ) = 0
∴ x + 5 + \(\sqrt{3}\) = 0 or x + 5 − \(\sqrt{3}\) = 0
Step 6 : Find the value of x
∴ x = − 5 − \(\sqrt{3}\) or x = − 5 + \(\sqrt{3}\)
∴ − 5 − \(\sqrt{3}\) , − 5 + \(\sqrt{3}\) are the roots of the quadratic equation
Recall formulae :
(a + b)^{2} = a^{2} + 2ab + b^{2} (a^{2} – b^{2}) = (a + b)(a – b) 
Examples :
(1) Solve : 5x^{2} − 4x − 3 = 0
It is convenient to make coefficient of x^{2} as 1 and then convert the equation as the of difference of two squares, so dividing the equation by 5,
we get, x^{2} – \(\frac{4}{5}\)x − \(\frac{3}{5}\) = 0
now if x^{2} – \(\frac{4}{5}\)x + k = (x − a)^{2} then x^{2} – \(\frac{4}{5}\)x + k = x^{2} − 2ax + a^{2}.
compare the terms
− 2ax = – \(\frac{4}{5}\)x, ∴ a = \(\frac{2}{5}\)
∴ k = a^{2} = \((\frac{2}{5})^2\) = \(\frac{4}{25}\)
Now, x^{2} – \(\frac{4}{5}\)x − \(\frac{3}{5}\) = 0
x^{2} – \(\frac{4}{5}\)x + \(\frac{4}{25}\) − \(\frac{4}{25}\) − \(\frac{3}{5}\) = 0
∴ \((x −\frac{2}{5})^2(\frac{4}{5}+\frac{3}{5})\) = 0
∴ \((x −\frac{2}{5})^2(\frac{19}{25})\) = 0
∴ \((x −\frac{2}{5})^2=(\frac{19}{25})\)
∴ \(x −\frac{2}{5}=\frac{\sqrt{19}}{5}\) or \(x −\frac{2}{5}=\frac{\sqrt{19}}{5}\)
∴ x = \(\frac{2}{5}+\frac{\sqrt{19}}{5}\) or x = \(\frac{2}{5}\frac{\sqrt{19}}{5}\)
∴ x = \(\frac{2+\sqrt{19}}{5}\) or x = \(\frac{2\sqrt{19}}{5}\)
∴ \(\frac{2+\sqrt{19}}{5}\) and \(\frac{2\sqrt{19}}{5}\) are roots of the equation.
(2) Solve : x^{2} + 8x − 48 = 0
Method I : Completing the square.
x^{2} + 8x − 48 = 0 ∴ x^{2} + 8x + 16 − 16 − 48 = 0 ∴ (x + 4)^{2} − 64 = 0 ∴ (x + 4)^{2} = 64 ∴ x + 4 = 8 or x + 4 = −8 ∴ x = 4 or x = − 12 
Method II : Factorisation
x^{2} + 8x − 48 = 0 ∴ x^{2} + 12x − 4x − 48 = 0 ∴ x (x + 12) − 4(x + 12) = 0 ∴ (x + 12) (x − 4) = 0 ∴ x + 12 = 0 or x − 4 = 0 ∴ x = −12 or x = 4 
Formula for solving a quadratic equation :
ax^{2} + bx + c, Divide the polynomial by a ( ‘.’ a ≠ 0) to get x^{2} + \(\frac{b}{a}\)x + \(\frac{c}{a}\).
Let us write the polynomial x^{2} + \(\frac{b}{a}\)x + \(\frac{c}{a}\) in the form of difference of two square numbers. Now we can obtain roots or solutions of equation x^{2} + x + = 0 which is equivalent to ax^{2} + bx + c = 0 .
ax^{2} + bx + c = 0
x^{2} + \(\frac{b}{a}\)x + \(\frac{c}{a}\) .....dividing both sides by a
∴ x^{2} + \(\frac{b}{a}\)x + \((\frac{b}{2a})^2\) − \((\frac{b}{2a})^2\) + \(\frac{c}{a}\) = 0
∴ \((x +\frac{b}{2a})^2\) − \(\frac{b^2}{4a^2}\) + \(\frac{c}{a}\) = 0
∴ \((x +\frac{b}{2a})^2\) − \(\frac{b^24ac}{4a^2}\) = 0
∴ \((x +\frac{b}{2a})^2\) = \(\frac{b^24ac}{4a^2}\)
∴ \(x +\frac{b}{2a}\) = \(\sqrt{\frac{b^24ac}{4a^2}}\) or \(x +\frac{b}{2a}\) = \(\sqrt{\frac{b^24ac}{4a^2}}\)
∴ \(x=\frac{b}{2a}+\sqrt{\frac{b^24ac}{4a^2}}\) or \(x=\frac{b}{2a}\sqrt{\frac{b^24ac}{4a^2}}\)
∴ \(x=\frac{b+\sqrt{b^24ac}}{2a}\) or \(x=\frac{b\sqrt{b^24ac}}{2a}\)
In short the solution is written as \(x=\frac{b±\sqrt{b^24ac}}{2a}\)
and these values are denoted by α, β.
∴ \(α=\frac{b+\sqrt{b^24ac}}{2a}\) , \(β=\frac{b\sqrt{b^24ac}}{2a}\) .....(1)
The values of a, b, c from equation ax^{2} + bx + c = 0 are substituted in \(\frac{b±\sqrt{b^24ac}}{2a}\) and further simplified to obtain the roots of the equation.
So \(x=\frac{b±\sqrt{b^24ac}}{2a}\) is the formula used to solve quadratic equation. Out of the two
roots any one can be represented by α and the other by β.
That is, instead (I) we can write
\(α=\frac{b+\sqrt{b^24ac}}{2a}\) , \(β=\frac{b\sqrt{b^24ac}}{2a}\) .....(2)
Note that : If \(α=\frac{b+\sqrt{b^24ac}}{2a}\) then α > β
If \(α=\frac{b\sqrt{b^24ac}}{2a}\) then α < β
For more information :
Solve the quadratic equation x^{2} − 2x −3 = 0 graphically x^{2} − 2x −3 = 0 ∴ x^{2} = 2x + 3 Let y = x^{2} = 2x + 3 . Let us draw graph of y = x^{2} and y = 2x + 3 y = x^{2}
y = 2x + 3
The graph of y = x^{2} is a parabola. The graphs of y = x^{2} and y = 2x + 3 intersect each other at (−1, 1) and (3, 9). ∴ x = −1 or x = 3 is the solution of the given quadratic equation. 
Solve the equation 2x^{2} + 13x + 15 = 0 by
(i) factorisation method
(ii) completing the square method
(iii) using formula. '
Verify that you will get the same roots every time.
(i) By factorisation method :
2x^{2} + 13x + 15 = 0
∴ 2x^{2} + 10x + 3x + 15 = 0
∴ 2x(x + 5) + 3(x + 5) = 0
∴ (x + 5)(2x + 3) = 0
∴ x + 5 = 0 or 2x + 3 = 0
∴ x = −5 or 2x = −3 ∴ x = \(\frac{3}{2}\)
∴ (— 5, \(\frac{3}{2}\) ) are the roots of the given quadratic equation.
(ii) By completing the square method
2x^{2} + 13x + 15 = 0
∴ x^{2} + \(\frac{13}{2}\)x + \(\frac{15}{2}\) = 0 …..(Dividing by 2)
now if x^{2} + x + k = (x + a)^{2} then x^{2} + \(\frac{13}{2}\)x + k = x^{2} + 2ax + a^{2}.
Comparing, \(\frac{13}{2}\) = 2a and k = a^{2}
∴ a = \(\frac{13}{4}\), k = \((\frac{13}{4})^2\) = \(\frac{169}{16}\)
x^{2} + \(\frac{13}{2}\)x + \(\frac{169}{16}\) — \(\frac{169}{16}\) + \(\frac{15}{2}\) = 0
∴ \((x+\frac{13}{4})^2(\frac{169120}{16})\) = 0
∴ \((x+\frac{13}{4})^2(\frac{49}{16})\) = 0
∴ \((x+\frac{13}{4})^2(\frac{7}{4})^2\) = 0
∴ \((x+\frac{13}{4}+\frac{7}{4})\)\((x+\frac{13}{4}\frac{7}{4})\) = 0
∴ (x + 5)(x + \(\frac{3}{2}\)) = 0
∴ x + 5 = 0 or x + \(\frac{3}{2}\) = 0
∴ x = −5 or x = \(\frac{3}{2}\)
∴ (— 5, \(\frac{3}{2}\)) are the roots of the given quadratic equation.
(iii) By using formula :
2x^{2} + 13x + 15 = 0
Comparing with ax^{2} + bx + c = 0
a = 2, b = 13, c = 15
b^{2} − 4ac = (13)^{2} − 4(2)(15)
= 169 — 120
= 49
x = \(x=\frac{b±\sqrt{b^24ac}}{2a}\)
∴ x = \(\frac{13±\sqrt{49}}{4}\) = \(\frac{13±7}{4}\)
∴ x = \(\frac{13+7}{4}\) or x = \(\frac{137}{4}\)
∴ x = \(\frac{6}{4}=\frac{3}{2}\) or x = \(\frac{20}{4}\) = −5
(— 5, \(\frac{3}{2}\)) are the roots of the given quadratic equation.
We get the same roots by all the three methods.
Nature of roots of a quadratic equation :
The roots of quadratic equation ax^{2} + bx + c = 0 are \(x=\frac{b±\sqrt{b^24ac}}{2a}\)
(1) If b^{2}− 4ac = 0 then, \(x=\frac{b±\sqrt{0}}{2a}\)
∴ \(x=\frac{b+0}{2a}\) or \(x=\frac{b0}{2a}\)
∴ the roots of the quadratic equation are real and equal.
(1) If b^{2} − 4ac > 0 then, \(x=\frac{b±\sqrt{b^24ac}}{2a}\)
∴ \(x=\frac{b+\sqrt{b^24ac}}{2a}\) , \(x=\frac{b\sqrt{b^24ac}}{2a}\)
∴ the roots of the quadratic equation are real and unequal.
(1) If b^{2} − 4ac < 0 then, \(x=\frac{b±\sqrt{b^24ac}}{2a}\) are not real numbers
∴ the roots of quadratic equations are not real.
[b^{2} − 4ac < 0 the value of b^{2} − 4ac is negative. The root of a negative number is not real.]
 The roots of the quadratic equation ax^{2} + bx + c = 0 are determined by the value of b^{2} − 4ac.
 Hence, b^{2} − 4ac is called the discriminant of the quadratic equation. It is denoted by Δ (Delta, a Greek letter).
Example :
(i) Fill in the blanks.
Value of discriminant  Nature of roots 
50  real and unequal 
−30  not real 
0  real and equal 
(ii) Determine nature of roots of the quadratic equations. 2x^{2} − 5x + 7 = 0
Solution : Compare 2x^{2} − 5x + 7 = 0 with
ax^{2} + bx + c = 0
a = 2, b = −5 , c = 7 ,
∴ b^{2} − 4 ac = (−5)^{2} − 4 × 2 × 7
D = 25 − 56
D = −31
∴ b^{2} − 4 ac < 0
∴ the roots of the equation are not real.
The relation between roots of the quadratic equation and coefficients :
The roots of quadratic equation ax^{2} + bx + c = 0 are α and β then.
\(α=\frac{b+\sqrt{b^24ac}}{2a}\) and \(β=\frac{b\sqrt{b^24ac}}{2a}\) .....(1)
α + β = \(\frac{b+\sqrt{b^24ac}}{2a}\)+\(\frac{b\sqrt{b^24ac}}{2a}\) = \(\frac{2b}{2a}\)
= \(\frac{b}{a}\)
The sum of the roots = \(\frac{b}{a}\) = \(\frac{coefficient\,\, of\,\,x}{coefficient\,\, of\,\,x^2}\)
α × β = \(\frac{b+\sqrt{b^24ac}}{2a}\)×\(\frac{b\sqrt{b^24ac}}{2a}\)
= \(\frac{(b)^2(\sqrt{b^24ac})^2}{4a^2}\)
= \(\frac{b^2b^2+4ac}{4a^2}\)
= \(\frac{c}{a}\)
The product of the roots = \(\frac{c}{a}\) = \(\frac{constant\,\,term}{coefficient\,\, of\,\,x^2}\)
Note : If α = 1 α + β = −b and α × β = c
Example :
Find α + β and α × β for 10x^{2} + 10x + 1 = 0,
Answer : α + β = \(\frac{b}{a}=\frac{10}{10}\) = −1
and α × β = \(\frac{c}{a}=\frac{1}{10}\)
To obtain a quadratic equation having given roots
If α and β be the roots of a quadratic equation in variable x then x = α and x = β
∴ x  α = 0 or x  β = 0
∴ (x  α)(x  β) = 0
∴ x^{2}  (α + β) x + α β = 0
When two roots of equation are given then quadratic equation can be obtained as
x^{2}  (addition of roots)x + product of the roots = 0.
Example :
(1) Write the quadratic equation if addition of the roots is 10 and product of the roots = 9
Answer :
Quadratic equation : x^{2}  (addition of roots)x + product of the roots = 0.
= x^{2}  (10)x + 9 = 0.
(2) What will be the quadratic equation if α = 2, β = 5
Solution :
It can be written as x^{2}  (α + β) x + α β = 0
∴ x^{2}  (2 + 5) x + 2×5 = 0
∴ x^{2}  7x + 10 = 0
(3) Obtain the quadratic equation if roots are 3, 7.
Solution : Let α = 3 and β = 7
∴ α + β = (3) + (7) = 10 and α × β = (3) × (7) = 21
∴ and quadratic equation is, x^{2}  (α + β) x + α β = 0
∴ x^{2} (10) x + 21 = 0
∴ x^{2 }+ 10x + 21 = 0
Remember :
(1) The roots of quadratic equation ax^{2} + bx + c = 0 are α and β then.
(2) The roots of the quadratic equation ax^{2} + bx + c = 0 are determined by the value of b^{2} − 4ac. Hence, b^{2} − 4ac is called the discriminant of the quadratic equation. It is denoted by Δ (Delta, a Greek letter). (3) If Δ = 0 The roots of quadratic equation are real and equal. If Δ > 0 then the roots of quadratic equation are real and unequal. If Δ < 0 then the roots of quadratic equation are not real. (4) If α and β be the roots of a quadratic equation then x^{2}  (α + β) x + α β = 0 
Application of quadratic equation :
Quadratic equations are useful to solve problems arising in our daytoday life.
The method of solving problems consists of the following three steps :
 Step 1 : Convert the word problem, into symbolic language, i.e. form mathematical equation by identifying the relationship existing in the problem.
 Step 2 : Solve the quadratic equation thus formed.
 Step 3 : Interpret the solution of the equation into verbal language. The appropriate solution/ solutions satisfying the given conditions is/ are to be considered.
Example :
(1) There is a rectangular onion storehouse in the farm of Mr. Ratnakarrao at Tivasa. The length of rectangular base is more than its breadth by 7 m and diagonal is more than length by 1 m. Find length and breadth of the storehouse.
Let breadth of the storehouse be x m.
∴ length = (x + 7) m, diagonal = x + 7 + 1 = (x + 8) m
By Pythagorus theorem
x^{2} + (x + 7)^{2} = (x + 8)^{2}
x^{2} + x^{2} + 14x + 49 = x^{2} + 16x + 64
∴ x^{2} + 14x  16x + 49  64 = 0
∴ x^{2}  2x 15 = 0
∴ x^{2}  5x + 3x  15 = 0
∴ x(x  5) + 3(x  5) = 0
∴ (x  5) (x + 3) = 0
∴ x  5 = 0 or x + 3 = 0
∴ x = 5 or x = 3
But length is never negative ∴ x ≠ 3
∴ x = 5 and x + 7 = 5 + 7 = 12
∴ Length of the base of storehouse is 12m and breadth is 5m.
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PDF : Class 10thMathematicsChapter2Quadratic EquationsNotes
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