Quadratic Equations Examples Archives

Solving A Quadratic Equation By Factoring

Since, 3x² – 5x + 2 is a quadratic polynomial;
3x² – 5x + 2 = 0 is a quadratic equation.
Also,
3x² – 5x + 2 = 3x² – 3x – 2x + 2 [Factorising]
= 3x (x – 1) – 2(x – 1)
= (x – 1) (3x – 2)
In the same way :
3x² – 5x + 2 = 0 ⇒ 3x² – 3x – 2x + 2 = 0 [Factorising L.H.S.]
⇒ (x – 1) (3x – 2) = 0
i.e., x – 1 = 0 or 3x – 2 = 0
⇒ x = 1 or x = 2/3
which is the solution of given quadratic equation.
In order to solve the given Quadratic Equation:
1. Clear the fractions and brackets, if given.
2. By transfering each term to the left hand side; express the given equation as
ax²+ bx + c = 0 or a + bx + cx² = 0
3. Factorise left hand side of the equation obtained (the right hand side being zero).
4. By putting each factor equal to zero; solve it.

Solving A Quadratic Equation By Factoring With Examples

Example 1: Solve (i) x² + 3x – 18 = 0 (ii) (x – 4) (5x + 2) = 0
(iii) 2x² + ax – a2 = 0; where ‘a’ is a real number.
Sol. (i) x² + 3x – 18 = 0
⇒ x² + 6x – 3x – 18 = 0
⇒ x(x + 6) – 3(x + 6) = 0
i.e., (x + 6) (x – 3) = 0
⇒ x + 6 = 0 or x – 3 = 0
⇒ x = – 6 or x = 3
Roots of the given equation are – 6 and 3
(ii) (x – 4) (5x + 2) = 0
⇒ x – 4 = 0 or 5x + 2 = 0
x = 4 or x = – 2/5
(iii) 2x² + ax – a² = 0
⇒ 2x²+ 2ax – ax – a²= 0
⇒ 2x(x + a) – a(x + a) = 0
i.e., (x + a) (2x – a) = 0
⇒ x + a = 0 or 2x – a = 0
⇒ x = – a or x = a/2

Example 2: Solve the following quadratic equations
(i) x² + 5x = 0 (ii) x² = 3x (iii) x² = 4
Sol. (i) x² + 5x = 0 ⇒ x(x + 5) = 0
⇒ x = 0 or x + 5 = 0
⇒ x = 0 or x = – 5
(ii) x² = 3x
⇒ x² – 3x = 0
⇒ x(x – 3) = 0
⇒ x = 0 or x = 3
(iii) x² = 4
⇒ x = ± 2

Example 3: Solve the following quadratic equations
(i) 7x² = 8 – 10x (ii) 3(x² – 4) = 5x (iii) x(x + 1) + (x + 2) (x + 3) = 42
Sol. (i) 7x² = 8 – 10x
⇒ 7x² + 10x – 8 = 0
⇒ 7x² + 14x – 4x – 8 = 0
⇒ 7x(x + 2) – 4(x + 2) = 0
⇒ (x + 2) (7x – 4) = 0
⇒ x + 2 = 0 or 7x – 4 = 0
⇒ x = – 2 or x = 4/7
(ii) 3(x² – 4) = 5x
⇒ 3x² – 5x – 12 = 0
⇒ 3x² – 9x + 4x –¬ 12 = 0
⇒ 3x(x – 3) + 4(x – 3) = 0
⇒ (x – 3) (3x + 4) = 0
⇒ x – 3 = 0 or 3x + 4 = 0
⇒ x = 3 or x = –4/3
(iii) x(x + 1) + (x + 2) (x + 3) = 42
⇒ x² + x + x² + 3x + 2x + 6 – 42 = 0
⇒ 2x² + 6x – 36 = 0
⇒ x² + 3x – 18 = 0
⇒ x² + 6x – 3x – 18 = 0
⇒ x(x + 6) – 3(x + 6) = 0
⇒ (x + 6) (x – 3) = 0
⇒ x = – 6 or x = 3

Example 4: Solve for x : 12abx² – (9a² – 8b²) x – 6ab = 0
Given equation is 12abx² – (9a² – 8b²) x – 6ab = 0
⇒ 3ax(4bx – 3a) + 2b(4bx – 3a) = 0
⇒ (4bx – 3a) (3ax + 2b) = 0
⇒ 4bx – 3a = 0 or 3ax + 2b = 0
⇒ x =3a/4b or x = – 2b/3a

Solving A Quadratic Equation By Completing The Square

Every quadratic equation can be converted in the form:
(x + a)² – b² = 0 or (x – a)² – b² = 0.
Steps:
1. Bring, if required, all the term of the quadratic equation to the left hand side.
2. Express the terms containing x as x² + 2xy or x² – 2xy.
3. Add and subtract y2 to get x² + 2xy + y² – y² or x² – 2xy + y² – y²; which gives
(x + y)² – y² or (x – y)² – y².
Thus,
(i) x² + 8x = 0 ⇒ x² + 2x × 4 = 0
⇒ x² + 2x × 4 + 4² – 4² = 0
⇒ (x + 4)² – 16 = 0
(ii) x² – 8x = 0 ⇒ x² – 2 × x × 4 = 0
⇒ x² – 2 × x × 4 + 4² – 4² = 0
⇒ (x – 4)² – 16 = 0

Solving A Quadratic Equation By Completing The Square 1

Solving A Quadratic Equation By Completing The Square With Examples

Example 1: Find the roots of the quadratic equation 2x² – 7x + 3 = 0
(if they exist) by the method of completing the square.
Sol.   2x² – 7x + 3 = 0
\(\Rightarrow {{x}^{2}}-\frac{7}{2}x+\frac{3}{2}=0 \)    [Dividing each term by 2]
\(\Rightarrow {{x}^{2}}-2\times x\times \frac{7}{4}+\frac{3}{2}=0 \)
\(\Rightarrow {{x}^{2}}-2\times x\times \frac{7}{4}+{{\left( \frac{7}{4} \right)}^{2}}-{{\left( \frac{7}{4} \right)}^{2}}+\frac{3}{2}=0 \)
\(\Rightarrow {{\left( x-\frac{7}{4} \right)}^{2}}-\frac{49}{16}+\frac{3}{2}=0 \)
\(\Rightarrow {{\left( x-\frac{7}{4} \right)}^{2}}-\left( \frac{49-24}{16} \right)=0 \)
\(\Rightarrow {{\left( x-\frac{7}{4} \right)}^{2}}-\frac{25}{16}=0 \)
\(i.e.,{{\left( x-\frac{7}{4} \right)}^{2}}=\frac{25}{16}\Rightarrow x-\frac{7}{4}=\pm \frac{5}{4} \)
\(i.e.,x-\frac{7}{4}=\frac{5}{4}\text{   or    }x-\frac{7}{4}=-\frac{5}{4}\text{ } \)
\(\Rightarrow x=\frac{7}{4}+\frac{5}{4}\text{   or    }x=\frac{7}{4}-\frac{5}{4}\text{ } \)
\(\Rightarrow x=3\text{ or    }x=\frac{1}{2} \)

Example 2: Find the roots of the quadratic equation 4x² + 4√3x + 3 = 0
Sol. 4x² + 4√3x + 3 = 0
\( \Rightarrow {{x}^{2}}+\sqrt{3}x+\frac{3}{4}=0 \)
\(i.e.,\text{ }{{x}^{2}}+2\times x\times \frac{\sqrt{3}}{2}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}-{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}+\frac{3}{4}=0 \)
\(\Rightarrow {{\left( x+\frac{\sqrt{3}}{2} \right)}^{2}}-\frac{3}{4}+\frac{3}{4}=0 \)
\(i.e.,{{\left( x+\frac{\sqrt{3}}{2} \right)}^{2}}=0 \)
\(\Rightarrow x+\frac{\sqrt{3}}{2}=0\text{ and }x=\frac{-\sqrt{3}}{2} \)

Example 3: Find the roots of the quadratic equation 2x² + x + 4 = 0
Sol. 2x² + x + 4 = 0
\(\Rightarrow {{x}^{2}}+\frac{x}{2}+2=0 \)
\(i.e.,~{{x}^{2}}+2\times x\times \frac{1}{4}+{{\left( \frac{1}{4} \right)}^{2}}-{{\left( \frac{1}{4} \right)}^{2}}+2=0 \)
\(\Rightarrow {{\left( x+\frac{1}{4} \right)}^{2}}-\frac{1}{16}+2=0 \)
\(\Rightarrow {{\left( x+\frac{1}{4} \right)}^{2}}+\frac{31}{16}=0 \)
\(i.e.,{{\left( x+\frac{1}{4} \right)}^{2}}=-\frac{31}{16} \)
This is not possible as the square of a real number can not be negative.

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