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Definition Of Quadratic Equation

 Any equation containing one term in which the unknown is squared and no term in which it is raised to a higher powersolve for x in the quadratic equation x2 + 4x + 4 = 0

Quadratic Equation Examples

Some examples of quadratic equations  (ax² + bx + c = 0):

• 6x² + 11x - 35 = 0
• 2x² - 4x - 2 = 0
• -4x² - 7x +12 = 0
• 20x² -15x - 10 = 0
• x² -x - 3 = 0
• 5x² - 2x - 9 = 0
• 3x² + 4x + 2 = 0
• -x² +6x + 18 = 0

Some examples of quadratic equations lacking the linear coefficient or the "bx":

• 2x² - 64 = 0
• x² - 16 = 0
• 9x² + 49 = 0
• -2x² - 4 = 0
• 4x² + 81 = 0
• -x² - 9 = 0
• 3x² - 36 = 0
• 6x² + 144 = 0

Other examples of quadratic equations lacking the constant term or "c":

• x² - 7x = 0
• 2x² + 8x = 0
• -x² - 9x = 0
• x² + 2x = 0
• -6x² - 3x = 0
• -5x² + x = 0
• -12x² + 13x = 0
• 11x² - 27x = 0

More examples of quadratic equation in factored form:

• (x + 2)(x - 3) = 0 [Gets on computing x² -1x - 6 = 0]
• (x + 1)(x + 6) = 0 [Gets on computing x² + 7x + 6 = 0]
• (x - 6)(x + 1) = 0 [Gets on computing x² - 5x - 6 = 0
• -3(x - 4)(2x + 3) = 0 [Gets on computing -6x² + 15x + 36 = 0]
• (x − 5)(x + 3) = 0 [Gets on computing x² − 2x − 15 = 0]
• (x - 5)(x + 2) = 0 [Gets on computing x² - 3x - 10 = 0]
• (x - 4)(x + 2) = 0 [Gets on computing x² - 2x - 8 = 0]
• (2x+3)(3x - 2) = 0 [Gets on computing 6x² + 5x - 6]

More examples of other forms of quadratic equations:

• x(x - 2) = 4 [upon Multiplicationand moving the 4 becomes x² - 2x - 4 = 0]
• x(2x + 3) = 12 [upon Multiplication and moving the 12 becomes 2x² - 3x - 12 = 0]
• 3x(x + 8) = -2 [upon Multiplication and moving the -2 becomes 3x² + 24x + 2 = 0]
• 5x² = 9 - x [moving the 9 and -x to the other side becomes 5x² + x - 9]
• -6x² = -2 + x [moving the -2 and x to the other side becomes -6x² - x + 2]
• x² = 27x -14 [moving the -14 and 27x to the other side becomes x² - 27x + 14]
• x² + 2x = 1 [moving "1" to the other side becomes x² + 2x - 1 = 0]
• 4x² - 7x = 15 [moving 15 to the other side becomes 4x² + 7x - 15 = 0]
• -8x² + 3x = -100 [moving -100 to the other side becomes -8x² + 3x + 100 = 0]
• 25x + 6 = 99 x² [moving 99 x2 to the other side becomes -99 x² + 25x + 6 = 0]

Solved examples of Quadratic equations

Lets solve

Problem 1: Solve for x: x²-3x-10 = 0

Solution:

Lets express -3x as a sum of -5x and +2x.

→ x²-5x+2x-10 = 0

→ x(x-5)+2(x-5) = 0

→ (x-5)(x+2) = 0

→ x-5 = 0        or         x+2 = 0

→ x = 5           or         x = -2

Problem 2: Solve for x: x²-18x+45 = 0

Solution:

The numbers which add up to -18 and give +45 when multiplied are -15 and -3.

Rewriting the equation,

→ x²-15x-3x+45 = 0

→ x(x-15)-3(x-15) = 0

→ (x-15) (x-3) = 0

→ x-15 = 0      or         x-3 = 0

→ x = 15         or         x = 3

Till now, the coefficient of x² was 1. 

Let us see how to solve the equations

 where the coefficient of x² is greater than 1.

Problem 3: Solve for x: 3x²+2x =1

Solution:

Rewrite our equation, we get 3x²+2x-1= 0

Here, the coefficient of x² is 3.

 In these cases, we multiply the constant c with the coefficient of x².

 Therefore, the product of the numbers we choose should be equal to -3 (-1*3).

Expressing 2x as a sum of +3x and –x

→ 3x²+3x-x-1 = 0

→ 3x(x+1)-1(x+1) = 0

→ (3x-1)(x+1) = 0

→ 3x-1 = 0      or         x+1 = 0

→ x = 1/3        or         x = -1

Problem 4: Solve for x: 11x²+18x+7 = 0

Solution:

In this case, the sum of the numbers we choose should equal to 18 and the product of the numbers should equal 11*7 = 77.

This can be done by expressing 18x as the sum of 11x and 7x.

→ 11x²+11x+7x+7 = 0

→ 11x(x+1) +7(x+1) = 0

→ (x+1)(11x+7) = 0

→ x+1 = 0       or         11x+7 = 0

→ x = -1          or         x = -7/11.

The factoring method is an easy way of finding the roots. But this method can be applied only to equations that can be factored.

For example, consider the equation x²+2x-6=0.

If we take +3 and -2, multiplying them gives -6 but adding them doesn’t give +2. Hence this quadratic equation cannot be factored.

For this kind of equations, we apply the quadratic formula to find the roots.

The quadratic formula to find the roots,

x = [-b ± √(b²-4ac)] / 2a

Now, let us find the roots of the equation above.

x²+2x-6 = 0

Here, a = 1, b=2 and c= -6.

Substituting these values in the formula,

x = [-2 ± √(4 – (4*1*-6))] / 2*1

→ x = [-2 ± √(4+24)] / 2

→ x = [-2 ± √28] / 2

When we get a non-perfect square in a square root, we usually try to express it as a product of two numbers in which one is a perfect square. This is for simplification purpose. Here 28 can be expressed as a product of 4 and 7.

→ x = [-2 ± √(4*7)] / 2

→ x = [-2 ± 2√7] / 2

→ x = 2[ -1 ± √7] / 2

→ x = -1 ± √7

Hence, √7-1 and -√7-1 are the roots of this equation.

Let us consider another example.

Solve for x: x² = 24 – 10x

Solution:

Rewriting the equation into the standard quadratic form,

x² +10x-24 = 0

What are the two numbers which when added give +10 and when multiplied give -24? 12 and -2.

So this can be solved by the factoring method. But let’s solve it using the new method, applying the quadratic formula.

Here, a = 1, b = 10 and c = -24.

x = [-10 ± √(100 – 4*1*-24)] / 2*1

x = [-10 ± √(100-(-96))] / 2

x = [-10 ± √196] / 2

x = [-10 ± 14] / 2

x = 2 or x= -12 are the roots.

Quadratic Equation Formula

                      wherea≠0andb2−4ac≥0

for any quadratic equation like:

$\purpleD{a}x^2 + \goldD{b}x + \redD{c} = 0$

Q1.Solve: x2 + 2x + 1 = 0

Solution

Given that a=1, b=2, c=1, and
Discriminant = b2 − 4ac = 22 − 4×1×1 = 0
Use the quadratic formula,  x = (−2 ± √0)/2 = −2/2
Therefore, x = − 1

Quadratic Equation Solver 

Quadratic Equation Calculator

Definition of quadratic equation  in Hindi

एक द्विघात समीकरण दूसरी डिग्री का एक समीकरण है, जिसका अर्थ है कि इसमें कम से कम एक शब्द होता है जिसे चुकता किया जाता है। मानक रूप ax standard + bx + c = 0 है जिसमें a, b, और c स्थिरांक या संख्यात्मक गुणांक है, और x एक अज्ञात चर है। एक पूर्ण नियम यह है कि पहला स्थिर "ए" शून्य नहीं हो सकता

quadratic equation Examples in Hindi

प्रश्न संख्या (1) (ii) $2x^2+x-4=0$

हल है 

दिया गया है, $2x^2+x-4=0$

यह पता करने के लिए कि दिये गये द्विघात समीकरण के वास्तविक मूल अस्तित्व में हैं या नहीं, हमें $D=b^2-4ac$ का मान ज्ञात करना होगा 

दिये गये द्विघात समीकरण में, $a=2,b=1$ and $c=-4$

$\therefore D=b^2-4ac$

$\Rightarrow D=1^2-4\times 2\times (-4)$

$\Rightarrow D=1+32=33$

Since, $D>0$ thus two roots of the given equation exist

चूँकि, $D=b^2-4ac>0$ अत: दिये गये द्विघात समीकरण के दो वास्तविक मूल होंगे।

दिया गया द्विघात समीकरण है, $2x^2+x-4=0$

पूर्ण वर्ग बनाने के क्रम में दिये गये द्विघात समीकरण के दोनों तरफ से 2 उभयनिष्ठ लेने पर

$\Rightarrow 2(x^2+\frac{x}{2}-\frac{4}{2})=0$

$\Rightarrow x^2+\frac{1}{2}x-2=0$

$\Rightarrow x^2+2\left(\frac{1}{4}\right)x-2=0$

$\Rightarrow x^2+2\left(\frac{1}{4}\right)x=2$

उपरोक्त समीकरण के दोनों तरफ ${\left(\frac{1}{4}\right)}^2$ जोड़नने पर

$\Rightarrow x^2+2\left(\frac{1}{4}\right)x+{\left(\frac{1}{4}\right)}^2=2+{\left(\frac{1}{4}\right)}^2$

$\Rightarrow {(x+\frac{1}{4})}^2=2+\frac{1}{16}$

$\Rightarrow {(x+\frac{1}{4})}^2=\frac{32+1}{16}$

$\Rightarrow {(x+\frac{1}{4})}^2=\frac{33}{16}$

$\Rightarrow x+\frac{1}{4}=\sqrt{\frac{33}{16}}$

$\Rightarrow x+\frac{1}{4}=\pm \frac{\sqrt{33}}{4}$

$\Rightarrow x=\pm \frac{\sqrt{33}}{4}-\frac{1}{4}$

$\Rightarrow x=\frac{\sqrt{33}}{4}-\frac{1}{4}\text{or}-\frac{\sqrt{33}}{4}-\frac{1}{4}$

$\Rightarrow x=\frac{\sqrt{33}-1}{4}\text{or}\frac{-\sqrt{33}-1}{4}$

अत: दिये गये समीकरन के दो मूल, $\frac{\sqrt{33}-1}{4}$ तथा $\frac{-\sqrt{33}-1}{4}$ हैं।

प्रश्न संख्या (1) (iii)$4x^2+4\sqrt{3}x+3=0$

हल:

दिया गया है, $4x^2+4\sqrt{3}x+3=0$

यह पता करने के लिए कि दिये गये द्विघात समीकरण के वास्तविक मूल अस्तित्व में हैं या नहीं, हमें $D=b^2-4ac$ के मान की गणना करनी होगी

दिये गये समीकरण में, $a=4,b=4\sqrt{3}$ and $c=3$

$\therefore D={\left(4\sqrt{3}\right)}^2-4\times 4\times 3$

$\Rightarrow D=48-48=0$

चूँकि, D = 0, अत: दिये गये द्विघात समीकरण का केवल एक ही वास्तविक मूल होगा।

अब, $4x^2+4\sqrt{3}x+3=0$

दिये गये द्विघात समीकरण में 4 उभयनिष्ठ लेने पर

$\Rightarrow 4(x^2+\frac{4\sqrt{3}}{4}x+\frac{3}{4})=0$

$\Rightarrow x^2+\sqrt{3}x+\frac{3}{4}=0$

पूर्ण बनाने के क्रम में दिये गये द्विघात समीकरण के मध्य पद को 2 से गुणा तथा भाग करने पर

$\Rightarrow x^2+2\left(\frac{\sqrt{3}}{2}\right)x+\frac{3}{4}=0$

$\Rightarrow x^2+2\left(\frac{\sqrt{3}}{2}\right)x=-\frac{3}{4}$

पूर्ण बनाने के क्रम में उपरोक्त समीकरण के दोनों तरफ ${\left(\frac{\sqrt{3}}{2}\right)}^2$ जोड़ने पर हम पाते हैं,

$x^2+2\left(\frac{\sqrt{3}}{2}\right)x+{\left(\frac{\sqrt{3}}{2}\right)}^2=-\frac{3}{4}+{\left(\frac{\sqrt{3}}{2}\right)}^2$

$\Rightarrow {(x+\frac{\sqrt{3}}{2})}^2=-\frac{3}{4}+\frac{3}{4}$

$\Rightarrow {(x+\frac{\sqrt{3}}{2})}^2=0$

$\Rightarrow x+\frac{\sqrt{3}}{2}=0$

$\Rightarrow x=-\frac{\sqrt{3}}{2}$

अत: दिये गये समीकरण का मूल है $-\frac{\sqrt{3}}{2}$ उत्तर