Question
Solve the following equations by using the method of completing the square:
$7x^2 + 3x - 4 = 0$
$7x^2 + 3x - 4 = 0$
✓
Answer
$7x^2 + 3x - 4 = 0$
$\Rightarrow 49x^2 + 21x - 28 = 0$ (Multiplying both sides by $7$)
$\Rightarrow 49x^2 + 21x = 28$
$\Rightarrow(\text{7x})^2+2\times\text{7x}\times\frac{3}{2}+\Big(\frac{3}{2}\Big)^2\\=28+\Big(\frac{3}{2}\Big)^2$
$\Big[$Adding $\Big(\frac{3}2{}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{7x}+\frac{3}{2}\Big)^2$
$=29+\frac{9}{4}$
$=\frac{121}{4}=\Big(\frac{11}{2}\Big)^2$
$\Rightarrow\text{7x}-\frac{3}{2}=\pm\frac{11}{2}$ (Taking square root on both sides)
$\Rightarrow\text{7x}+\frac{3}{2}=\frac{11}{2}$ or $\text{7x}+\frac{3}{2}=-\frac{11}{2}$
$\Rightarrow\text{7x}=\frac{11}{2}-\frac{3}{2}=\frac{8}{2}=4$ or $\text{7x}=-\frac{11}{2}-\frac{3}{2}=-\frac{14}{2}=-7$
$\Rightarrow\text{x}=\frac{4}{7}$ or $x = -1$
Hence, $\frac{4}{7}$ and $-1$ are the roots of the given equation.
$\Rightarrow 49x^2 + 21x - 28 = 0$ (Multiplying both sides by $7$)
$\Rightarrow 49x^2 + 21x = 28$
$\Rightarrow(\text{7x})^2+2\times\text{7x}\times\frac{3}{2}+\Big(\frac{3}{2}\Big)^2\\=28+\Big(\frac{3}{2}\Big)^2$
$\Big[$Adding $\Big(\frac{3}2{}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{7x}+\frac{3}{2}\Big)^2$
$=29+\frac{9}{4}$
$=\frac{121}{4}=\Big(\frac{11}{2}\Big)^2$
$\Rightarrow\text{7x}-\frac{3}{2}=\pm\frac{11}{2}$ (Taking square root on both sides)
$\Rightarrow\text{7x}+\frac{3}{2}=\frac{11}{2}$ or $\text{7x}+\frac{3}{2}=-\frac{11}{2}$
$\Rightarrow\text{7x}=\frac{11}{2}-\frac{3}{2}=\frac{8}{2}=4$ or $\text{7x}=-\frac{11}{2}-\frac{3}{2}=-\frac{14}{2}=-7$
$\Rightarrow\text{x}=\frac{4}{7}$ or $x = -1$
Hence, $\frac{4}{7}$ and $-1$ are the roots of the given equation.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Solve the following quadratic equation:
$\frac{1}{\text{x}-1}-\frac{1}{\text{x}+5}=\frac{6}{7},$ $\text{x}\neq1,\ -5$
→$\frac{1}{\text{x}-1}-\frac{1}{\text{x}+5}=\frac{6}{7},$ $\text{x}\neq1,\ -5$
Draw a circle with centre O and radius 4cm. Draw any diameter AB of this circle. Construct tangents to the circle at each of the two end points of the diameter AB.
A ma saved $₹ 33000$ in $10$ month. In each month after the first, he saved $₹ 100$ more than he did in the preceding month. How much did he save in the first month?
→ABCD is a field in the shape of a trapezium, $\text{AD}||\text{BC},\ \angle\text{ABC}=90^\circ$ and $LADC 60^\circ .$ $\angle\text{ADC}=60^\circ.$ Four sectors are formed with centres $A, B, C$ and $D$, as shown in the figure. The radius of each sector is $14\ m$. Find the following.
→- Total area of the four sector,
- Area of the remaining portion, given that $AD =55m, BC = 45m$ and $AB = 30m,$
Find two numbers such that the sum of twice the first and thrice the second is 92, and four times the first exceeds seven times the second by 2
The sum of first 10 term of an AP is -150 and the sum of its next 10 terms is -550. Find the AP.
In a right $\triangle\text{ABC},$ right-angled at B, if $\tan\text{A}=1$ then verify that $2\sin\text{A}\cdot\cos\text{A}=1.$
→In the given figure, D is the midpoint of side BC and $\text{AE}\perp\text{BC}.$ If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that.
$\text{b}^2=\text{p}^2+\text{ax}+\frac{\text{a}^2}{4}$
→$\text{b}^2=\text{p}^2+\text{ax}+\frac{\text{a}^2}{4}$
Prove the following identities:
$\frac{\tan\text{A}+\tan\text{B}}{\cot\text{A}+\cot\text{B}}=\tan\text{A}\tan\text{B}$
→$\frac{\tan\text{A}+\tan\text{B}}{\cot\text{A}+\cot\text{B}}=\tan\text{A}\tan\text{B}$
Find the roots of the following equation, if they exist, by applying the quadratic formula:
$4x^2 + 4bx - (a^2 - b^2) = 0$
→$4x^2 + 4bx - (a^2 - b^2) = 0$