# Quadratic Equations | Completing Squares

Ex 4.3 Q1. NCERT Solutions For Class 10 Maths

#### Question: Find the roots of the following quadratic equations, if they exist, by the method of completing squares:

i. 2x2 - 7x + 3 = 0
ii. 2x2 + x – 4 = 0
iii. 4x2+ 4$$sqrt3$x + 3 = 0 iv. 2x2 + x + 4 = 0 #### Video Explanation for i. 2x2 - 7x + 3 = 0 #### Explanatory Answer ##### i. 2x2 - 7x + 3 = 0 | Method 1 Step 1: Divide the entire equation by 2 so that the coefficient of x2 is 1. x2 - $\frac{7}{2}$x + $\frac{3}{2}$ = 0 Step 2: Divide the coefficient of x by 2. i.e., $\frac{–7}{2}$ by 2 and complete the square as shown below.$x - $$frac {7} {4}{$}^{2}) – $$$frac{7}{4}{$}^{2}) + $\frac{3}{2}$ = 0$x - $$frac {7} {4}{$}^{2}) - $\frac{49}{16}$ + $\frac{3}{2}$ = 0$x - $$frac {7} {4}{$}^{2}) - $\frac{49 - 24}{16}$ = 0$x - $$frac {7} {4}{$}^{2}) = $\frac{25}{16}$ x - $\frac{7}{4}$ = ±$\frac{5}{4}$ x = $\frac{5}{4}$ + $\frac{7}{4}$ and x = -$\frac{5}{4}$ + $\frac{7}{4}$ x = 3 and x = $\frac{1}{2}$ ##### i. 2x2 - 7x + 3 = 0 | Method 2 Step 1: Multiply the entire equation by 2 so that the coefficient of x2 is a perfect square. 4x2 – 14x + 6 = 0 Step 2: We have to expres 4x2 – 14x + 6 = 0 as a square of the form$p - q)2 ± something = 0.
Let us find out what p and q are. 4x2 is p2
∴ p = 2x.
So, the square will be (2x - q)2
Only step left to complete the square is find q. Let us find that.
2pq = 14x
p = 2x. So, 4xq = 14x. Or q = $$frac{7}{2}$ So, the square is completed as shown below$2x - $$frac {7} {2}{$}^{2}) – $$$frac{7}{2}{$}^{2}) + 6 = 0$2x - $$frac {7} {2}{$}^{2}) – $\frac{49}{4}$ + 6 = 0$2x - $$frac {7} {2}{$}^{2}) - $\frac{49 - 24}{4}$ = 0$2x - $$frac {7} {2}{$}^{2}) – $\frac{25}{4}$ = 0$2x - $$frac {7} {2}{$}^{2}) = $\frac{25}{4}$$2x – $$frac{7}{2}$) = ±$\frac{5}{2}$ 2x = $\frac{7}{2}$ + $\frac{5}{2}$ and 2x = $\frac{7}{2}$ – $\frac{5}{2}$ 2x = 6 and 2x = 1 x = 3 and x = $\frac{1}{2}$ #### Video Explanation for ii, iii, and iv ##### ii. 2x2 + x – 4 = 0 | Method 1 Step 1: Divide the entire equation by 2 so that the coefficient of x2 is 1. x2 + $\frac{x}{2}$ – 2 = 0 Step 2: Divide the coefficient of x by 2. i.e., $\frac{1}{2}$ by 2 and complete the square as shown below.$x - $$frac {1} {4}{$}^{2}) – $$$frac{1}{4}{$}^{2}) – 2 = 0$x - $$frac {1} {4}{$}^{2}) – $\frac{1}{16}$ - 2 = 0$x - $$frac {1} {4}{$}^{2}) – $\frac{33}{16}$ = 0$x - $$frac {1} {4}{$}^{2}) = $\frac{33}{16}$ Or x + $\frac{1}{4}$ = ± $\frac{\sqrt{33}}{4}$ x = $$$frac{-1 + \sqrt{33}}{4}$) or x = $$$frac{-1 - \sqrt{33}}{4}$) ##### ii. 2x2 + x – 4 = 0 | Method 2 Step 1: Multiply the entire equation by 2 so that the coefficient of x2 is a perfect square. 4x2 + 2x – 8 = 0 Step 2: We have to expres 4x2 + 2x – 8 = 0 as a square of the form$p + q)2 ± something = 0.
Let us find out what p and q are. 4x2 is p2
∴ p = 2x.
So, the square will be (2x + q)2
Only step left to complete the square is find q. Let us find that.
2pq = 2x
p = 2x. So, 4xq = 2x. Or q = $$frac{1}{2}$ So, the square is completed as shown below$2x + $$frac {1} {2}{$}^{2}) – $$$frac{1}{2}{$}^{2}) – 8 = 0$2x + $$frac {1} {2}{$}^{2}) – $\frac{1}{4}$ - 8 = 0$2x + $$frac {1} {2}{$}^{2}) – $\frac{33}{4}$ = 0$2x + $$frac {1} {2}{$}^{2}) = $\frac{33}{4}$ $\frac{2x+1}{2}$ = ±$\frac{\sqrt33}{2}$ x = ±$\frac{{\sqrt33} - 1}{4}$ or x = $\frac{{-\sqrt33} - 1}{4}$ ##### iii. 4x2 + 4√3 x + 3 = 0 | Method 1 Step 1: Divide the entire equation by 4 so that the coefficient of x2 is 1. x2 + $\sqrt{3}$x + $\frac{3}{4}$ = 0 Step 2: Divide the coefficient of x by 2. i.e., $\sqrt3$ by 2 and complete the square as shown below.$x + $$frac {\sqrt3} {2}{$}^{2}) - $$$frac{\sqrt3}{2}{$}^{2}) + $\frac{3}{4}$ = 0$x + $$frac {\sqrt3} {2}{$}^{2}) -$\frac{3}{4}$ + $\frac{3}{4}$ = 0$x + $$frac {\sqrt3} {2}{$}^{2}) = 0 x = - $\frac{\sqrt3}{2}$ and -$\frac{\sqrt3}{2}$ ##### iii. 4x2 + 4√3 x + 3 = 0 | Method 2 Step 1: The coefficient of x2 is a perfect square. So, there is nothing to be multiplied. Step 2: We have to expres 4x2 + 4$\sqrt{3}$x + 3 = 0 as a square of the form$p + q)2 ± something = 0.
Let us find out what p and q are. 4x2 is p2
∴ p = 2x.
So, the square will be (2x + q)2
Only step left to complete the square is find q. Let us find that.
2pq = 4$$sqrt{3}$x p = 2x. So, 4xq = 4$\sqrt{3}$x. Or q = $\sqrt{3}$ So, the square is completed as shown below$2x + $$sqrt3$)2 -$$$sqrt3$)2 + 3 = 0$2x + $$sqrt3$)2 - 3 + 3 = 0$2x + $$sqrt3$)2 = 0 2x + $\sqrt3$ = 0 x = -$\frac{\sqrt3}{2}$ and - $\frac{\sqrt3}{2}$ ##### iv. 2x2+ x + 4 = 0 | Method 1 Step 1: Divide the entire equation by 2 so that the coefficient of x2 is 1. x2 + $\frac{x}{2}$ + 2 = 0 Step 2: Divide the coefficient of x by 2. i.e., $\frac{1}{2}$ by 2 and complete the square as shown below.$x + $$frac {1} {4}{$}^{2}) - $$$frac{1}{4}{$}^{2}) + 2 = 0$x + $$frac {1} {4}{$}^{2}) - $\frac{1}{16}$ + 2 = 0$x + $$frac {1} {4}{$}^{2}) + $\frac{32 – 1}{16}$ = 0$x + $$frac {1} {4}{$}^{2}) = -$\frac{31}{16}$ < 0$x + $$frac {1} {4}{$}^{2}) cannot be negative for any real value of x because square of real numbers is always non - negative. ∴ the given equation has no real roots. ##### iv. 2x2+ x + 4 = 0 | Method 2 Step 1: Multiply the entire equation by 2 so that the coefficient of x2 is a perfect square. 4x2 + 2x + 8 = 0 Step 2: We have to expres 4x2 + 2x + 8 = 0 as a square of the form$p + q)2 ± something = 0.
Let us find out what p and q are. 4x2 is p2
∴ p = 2x.
So, the square will be (2x + q)2
Only step left to complete the square is find q. Let us find that.
2pq = 2x
p = 2x. So, 4xq = 2x. Or q = $$frac{1}{2}$ So, the square is completed as shown below$2x + $$frac {1} {2}{$}^{2}) - $$$frac{1}{2}{$}^{2}) + 8 = 0$2x + $$frac {1} {2}{$}^{2}) - $\frac{1}{4}$ + 8 = 0$2x + $$frac {1} {2}{$}^{2}) + $\frac{32-1}{4}$ = 0$2x + $$frac {1} {2}{$}^{2}) = -$\frac{31}{4}$ < 0$2x + $\frac {1} {2}{$}^{2}) cannot be negative for any real value of x because square of real numbers is always non - negative.
∴ the given equation has no real roots.