SEBA Class 10 Maths Exercise 4.3 Solutions: Quadratic Equations | New Book 2026
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SEBA Class 10 Maths Exercise 4.1 Solutions
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SEBA Class 10 Maths Exercise 4.3 Solutions
SEBA Maths Exercise 4.3
1. Find the roots of the following quadratic equations applying quadratic formula:
(i) $2x^{2}-7x+3=0$
Soln .
Given: $2x^{2}-7x+3=0$
Here, $a=2$, $b=-7$, and $c=3$
So, $b^{2}-4ac$
$=(-7)^{2}-4\times2\times3$
$=49-24$
$=25>0$
Therefore:
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$=\frac{-(-7)\pm\sqrt{25}}{2\times2}$
$=\frac{7\pm5}{4}$
Either, $x=\frac{7+5}{4}$
$\Rightarrow x=\frac{12}{4}$
$\Rightarrow x=3$
or, $x=\frac{7-5}{4}$
$\Rightarrow x=\frac{2}{4}$
$\Rightarrow x=\frac{1}{2}$
So, the roots are 3 and $\frac{1}{2}$.
(ii) $2x^{2}+x-4=0$
Soln .
Given: $2x^{2}+x-4=0$
Here, $a=2$, $b=1$, and $c=-4$
So, $b^{2}-4ac$
$=(1)^{2}-4\times2\times(-4)$
$=1+32$
$=33>0$
Therefore:
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$=\frac{-1\pm\sqrt{33}}{2\times2}$
$=\frac{-1\pm\sqrt{33}}{4}$
Either, $x=\frac{-1+\sqrt{33}}{4}$
or, $x=\frac{-1-\sqrt{33}}{4}$
So, the roots are $\frac{-1+\sqrt{33}}{4}$ and $\frac{-1-\sqrt{33}}{4}$.
(iii) $4x^{2}+4\sqrt{3}x+3=0$
Soln .
Given: $4x^{2}+4\sqrt{3}x+3=0$
Here, $a=4$, $b=4\sqrt{3}$, and $c=3$
So, $b^{2}-4ac$
$=(4\sqrt{3})^{2}-4\times4\times3$
$=16\times3-48$
$=48-48$
$=0$
Therefore:
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$=\frac{-4\sqrt{3}\pm0}{2\times4}$
$=\frac{-4\sqrt{3}}{8}$
$=-\frac{\sqrt{3}}{2}$
So, the roots are $-\frac{\sqrt{3}}{2}$ and $-\frac{\sqrt{3}}{2}$.
(iv) $2x^{2}+5\sqrt{3}x+6=0$
Soln .
Given: $2x^{2}+5\sqrt{3}x+6=0$
Here, $a=2$, $b=5\sqrt{3}$, and $c=6$
So, $b^{2}-4ac$
$=(5\sqrt{3})^{2}-4\times2\times6$
$=25\times3-48$
$=75-48$
$=27>0$
Therefore:
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$=\frac{-5\sqrt{3}\pm\sqrt{27}}{2\times2}$
$=\frac{-5\sqrt{3}\pm3\sqrt{3}}{4}$
Either, $x=\frac{-5\sqrt{3}+3\sqrt{3}}{4}$
$\Rightarrow x=\frac{-2\sqrt{3}}{4}$
$\Rightarrow x=-\frac{\sqrt{3}}{2}$
or, $x=\frac{-5\sqrt{3}-3\sqrt{3}}{4}$
$\Rightarrow x=\frac{-8\sqrt{3}}{4}$
$\Rightarrow x=-2\sqrt{3}$
So, the roots are $-\frac{\sqrt{3}}{2}$ and $-2\sqrt{3}$.
(v) $x^{2}+4x+1=0$
Soln .
Given: $x^{2}+4x+1=0$
Here, $a=1$, $b=4$, and $c=1$
So, $b^{2}-4ac$
$=(4)^{2}-4\times1\times1$
$=16-4$
$=12>0$
Therefore:
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$=\frac{-4\pm\sqrt{12}}{2\times1}$
$=\frac{-4\pm2\sqrt{3}}{2}$
$=\frac{2(-2\pm\sqrt{3})}{2}$
$=-2\pm\sqrt{3}$
Either, $x=-2+\sqrt{3}$
or, $x=-2-\sqrt{3}$
So, the roots are $-2+\sqrt{3}$ and $-2-\sqrt{3}$.
(vi) $4x^{2}+x-3=0$
Soln .
Given: $4x^{2}+x-3=0$
Here, $a=4$, $b=1$, and $c=-3$
So, $b^{2}-4ac$
$=(1)^{2}-4\times4\times(-3)$
$=1+48$
$=49>0$
Therefore:
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$=\frac{-1\pm\sqrt{49}}{2\times4}$
$=\frac{-1\pm7}{8}$
Either, $x=\frac{-1+7}{8}$
$\Rightarrow x=\frac{6}{8}$
$\Rightarrow x=\frac{3}{4}$
or, $x=\frac{-1-7}{8}$
$\Rightarrow x=\frac{-8}{8}$
$\Rightarrow x=-1$
So, the roots are $\frac{3}{4}$ and $-1$.
(vii) $abx^{2}+(b^{2}-ac)x-bc=0$
Soln .
Given: $abx^{2}+(b^{2}-ac)x-bc=0$
Here, $a=ab$, $b=(b^{2}-ac)$, and $c=-bc$
So, $b^{2}-4ac$
$=(b^{2}-ac)^{2}-4\times(ab)\times(-bc)$
$=(b^{2}-ac)^{2}+4ab^{2}c$
$=(b^{2})^{2}-2b^{2}ac+(ac)^{2}+4ab^{2}c$
$=(b^{2})^{2}+2b^{2}ac+(ac)^{2}$
$=(b^{2}+ac)^{2}>0$
Therefore:
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$=\frac{-(b^{2}-ac)\pm\sqrt{(b^{2}+ac)^{2}}}{2ab}$
$=\frac{-(b^{2}-ac)\pm(b^{2}+ac)}{2ab}$
Either, $x=\frac{-(b^{2}-ac)+(b^{2}+ac)}{2ab}$
$\Rightarrow x=\frac{-b^{2}+ac+b^{2}+ac}{2ab}$
$\Rightarrow x=\frac{2ac}{2ab}$
$\Rightarrow x=\frac{c}{b}$
or, $x=\frac{-(b^{2}-ac)-(b^{2}+ac)}{2ab}$
$\Rightarrow x=\frac{-b^{2}+ac-b^{2}-ac}{2ab}$
$\Rightarrow x=\frac{-2b^{2}}{2ab}$
$\Rightarrow x=-\frac{b}{a}$
So, the roots are $\frac{c}{b}$ and $-\frac{b}{a}$.
(viii) $x^{2}-2ax+(a^{2}-b^{2})=0$
Soln .
Given: $x^{2}-2ax+(a^{2}-b^{2})=0$
Here, $a=1$, $b=-2a$, and $c=(a^{2}-b^{2})$
So, $b^{2}-4ac$
$=(-2a)^{2}-4\times1\times(a^{2}-b^{2})$
$=4a^{2}-4a^{2}+4b^{2}$
$=4b^{2}\ge0$
Therefore:
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$=\frac{-(-2a)\pm\sqrt{4b^{2}}}{2\times1}$
$=\frac{2a\pm2b}{2}$
$=\frac{2(a\pm b)}{2}$
$=a\pm b$
Either, $x=a+b$
or, $x=a-b$
So, the roots are $a+b$ and $a-b$.
(ix) $a(x^{2}+1)=x(a^{2}+1)$
Soln .
Given: $a(x^{2}+1)=x(a^{2}+1)$
$\Rightarrow ax^{2}+a=a^{2}x+x$
$\Rightarrow ax^{2}-(a^{2}+1)x+a=0$
Here, $a=a$, $b=-(a^{2}+1)$, and $c=a$
So, $b^{2}-4ac$
$=[-(a^{2}+1)]^{2}-4\times a\times a$
$=(a^{2}+1)^{2}-4a^{2}$
$=(a^{2})^{2}+2a^{2}+1-4a^{2}$
$=(a^{2})^{2}-2a^{2}+1$
$=(a^{2}-1)^{2}\ge0$
Therefore:
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$=\frac{-[-(a^{2}+1)]\pm\sqrt{(a^{2}-1)^{2}}}{2a}$
$=\frac{(a^{2}+1)\pm(a^{2}-1)}{2a}$
Either, $x=\frac{(a^{2}+1)+(a^{2}-1)}{2a}$
$\Rightarrow x=\frac{2a^{2}}{2a}$
$\Rightarrow x=a$
or, $x=\frac{(a^{2}+1)-(a^{2}-1)}{2a}$
$\Rightarrow x=\frac{a^{2}+1-a^{2}+1}{2a}$
$\Rightarrow x=\frac{2}{2a}$
$\Rightarrow x=\frac{1}{a}$
So, the roots are $a$ and $\frac{1}{a}$.
(x) $8x(2x-1)+1=0$
Soln .
Given: $8x(2x-1)+1=0$
$\Rightarrow 16x^{2}-8x+1=0$
Here, $a=16$, $b=-8$, and $c=1$
So, $b^{2}-4ac$
$=(-8)^{2}-4\times16\times1$
$=64-64$
$=0$
Therefore:
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$=\frac{-(-8)\pm0}{2\times16}$
$=\frac{8}{32}$
$=\frac{1}{4}$
So, the roots are $\frac{1}{4}$ and $\frac{1}{4}$.
2. Express the equation $\frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3}$, $x\ne3$, $x\ne5$ as a quadratic equation in standard form. Hence find the roots of the equation so formed.
Soln .
Given: $\frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3}$
$\Rightarrow \frac{(x-2)(x-5)+(x-4)(x-3)}{(x-3)(x-5)}=\frac{10}{3}$
$\Rightarrow \frac{(x^{2}-5x-2x+10)+(x^{2}-3x-4x+12)}{x^{2}-5x-3x+15}=\frac{10}{3}$
$\Rightarrow \frac{x^{2}-7x+10+x^{2}-7x+12}{x^{2}-8x+15}=\frac{10}{3}$
$\Rightarrow \frac{2x^{2}-14x+22}{x^{2}-8x+15}=\frac{10}{3}$
$\Rightarrow 3(2x^{2}-14x+22)=10(x^{2}-8x+15)$
$\Rightarrow 6x^{2}-42x+66=10x^{2}-80x+150$
$\Rightarrow 10x^{2}-6x^{2}-80x+42x+150-66=0$
$\Rightarrow 4x^{2}-38x+84=0$
$\Rightarrow (2x^{2}-19x+42)=0$
∴$2x^{2}-19x+42=0$
Here, $a=2$, $b=-19$, and $c=42$
So, the discriminant, $b^{2}-4ac$
$=(-19)^{2}-4\times2\times42$
$=361-336$
$=25>0$
Therefore:
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$=\frac{-(-19)\pm\sqrt{25}}{2\times2}$
$=\frac{19\pm5}{4}$
Either, $x=\frac{19+5}{4}$
$\Rightarrow x=\frac{24}{4}$
$\Rightarrow x=6$
or, $x=\frac{19-5}{4}$
$\Rightarrow x=\frac{14}{4}$
$\Rightarrow x=\frac{7}{2}$
So, the quadratic equation in standard form is $2x^{2}-19x+42=0$, and its roots are 6 and $\frac{7}{2}$.
3. Find the roots of the following equations:
(i) $x-\frac{1}{x}=3$, $x\ne0$
Soln .
Given: $x-\frac{1}{x}=3$
$\Rightarrow \frac{x^{2}-1}{x}=3$
$\Rightarrow x^{2}-1=3x$
$\Rightarrow x^{2}-3x-1=0$
Here, $a=1$, $b=-3$, and $c=-1$
So, the discriminant, $b^{2}-4ac$
$=(-3)^{2}-4\times1\times(-1)$
$=9+4$
$=13>0$
Therefore:
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$=\frac{-(-3)\pm\sqrt{13}}{2\times1}$
$=\frac{3\pm\sqrt{13}}{2}$
Either, $x=\frac{3+\sqrt{13}}{2}$
or, $x=\frac{3-\sqrt{13}}{2}$
So, the roots are $\frac{3+\sqrt{13}}{2}$ and $\frac{3-\sqrt{13}}{2}$.
(ii) $\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}$, $x\ne-4,7$
Soln .
Given: $\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}$
$\Rightarrow \frac{(x-7)-(x+4)}{(x+4)(x-7)}=\frac{11}{30}$
$\Rightarrow \frac{x-7-x-4}{x^{2}-7x+4x-28}=\frac{11}{30}$
$\Rightarrow \frac{-11}{x^{2}-3x-28}=\frac{11}{30}$
$\Rightarrow \frac{-1}{x^{2}-3x-28}=\frac{1}{30}$
$\Rightarrow x^{2}-3x-28=-30$
$\Rightarrow x^{2}-3x-28+30=0$
$\Rightarrow x^{2}-3x+2=0$
Here, $a=1$, $b=-3$, and $c=2$
So, $b^{2}-4ac$
$=(-3)^{2}-4\times1\times2$
$=9-8$
$=1>0$
Therefore:
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$=\frac{-(-3)\pm\sqrt{1}}{2\times1}$
$=\frac{3\pm1}{2}$
Either, $x=\frac{3+1}{2}$
$\Rightarrow x=\frac{4}{2}$
$\Rightarrow x=2$
or, $x=\frac{3-1}{2}$
$\Rightarrow x=\frac{2}{2}$
$\Rightarrow x=1$
So, the roots are 2 and 1.
(iii) $\frac{2}{3}x^{2}-\frac{1}{3}x-1=0$
Soln . Given:
$\frac{2}{3}x^{2}-\frac{1}{3}x-1=0$
Multiplying throughout by 3, we get:
$2x^{2}-x-3=0$
Here, $a=2$, $b=-1$, and $c=-3$
So, the discriminant, $b^{2}-4ac$
$=(-1)^{2}-4\times2\times(-3)$
$=1+24$
$=25>0$
Therefore:
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$=\frac{-(-1)\pm\sqrt{25}}{2\times2}$
$=\frac{1\pm5}{4}$
Either, $x=\frac{1+5}{4}$
$\Rightarrow x=\frac{6}{4}$
$\Rightarrow x=\frac{3}{2}$
or, $x=\frac{1-5}{4}$
$\Rightarrow x=\frac{-4}{4}$
$\Rightarrow x=-1$
So, the roots are $\frac{3}{2}$ and $-1$.
(iv) $2x^{2}+\frac{1}{2}=2x$
Soln .
Given: $2x^{2}+\frac{1}{2}=2x$
$\Rightarrow \frac{4x^{2}+1}{2}=2x$
$\Rightarrow 4x^{2}+1=4x$
$\Rightarrow 4x^{2}-4x+1=0$
Here, $a=4$, $b=-4$, and $c=1$
So, $b^{2}-4ac$
$=(-4)^{2}-4\times4\times1$
$=16-16$
$=0$
Therefore:
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$=\frac{-(-4)\pm0}{2\times4}$
$=\frac{4}{8}$
$=\frac{1}{2}$
So, the roots are $\frac{1}{2}$ and $\frac{1}{2}$.
(v) $x+\frac{1}{x}=2$
Soln .
Given: $x+\frac{1}{x}=2$
$\Rightarrow \frac{x^{2}+1}{x}=2$
$\Rightarrow x^{2}+1=2x$
$\Rightarrow x^{2}-2x+1=0$
Here, $a=1$, $b=-2$, and $c=1$
So, $b^{2}-4ac$
$=(-2)^{2}-4\times1\times1$
$=4-4$
$=0$
Therefore:
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$=\frac{-(-2)\pm0}{2\times1}$
$=\frac{2}{2}$
$=1$
So, the roots are 1 and 1.
(vi) $\frac{5x-6}{4x-1}=\frac{2x+3}{3x+2}$
Soln .
Given: $\frac{5x-6}{4x-1}=\frac{2x+3}{3x+2}$
$\Rightarrow (5x-6)(3x+2)=(4x-1)(2x+3)$
$\Rightarrow 15x^{2}+10x-18x-12=8x^{2}+12x-2x-3$
$\Rightarrow 15x^{2}-8x-12=8x^{2}+10x-3$
$\Rightarrow 15x^{2}-8x^{2}-8x-10x-12+3=0$
$\Rightarrow 7x^{2}-18x-9=0$
Here, $a=7$, $b=-18$, and $c=-9$
So, the discriminant, $b^{2}-4ac$
$=(-18)^{2}-4\times7\times(-9)$
$=324+252$
$=576>0$
Therefore:
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$=\frac{-(-18)\pm\sqrt{576}}{2\times7}$
$=\frac{18\pm24}{14}$
Either, $x=\frac{18+24}{14}$
$\Rightarrow x=\frac{42}{14}$
$\Rightarrow x=3$
or, $x=\frac{18-24}{14}$
$\Rightarrow x=\frac{-6}{14}$
$\Rightarrow x=-\frac{3}{7}$
So, the roots are 3 and $-\frac{3}{7}$.
4. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is $\frac{1}{3}$. Find his present age.
Soln .
Let the present age of Rehman be $x$ years.
3 years ago, Rehman’s age was $(x-3)$ years, and
5 years from now, Rehman’s age will be $(x+5)$ years.
According to the question:
$\frac{1}{x-3}+\frac{1}{x+5}=\frac{1}{3}$
$\Rightarrow \frac{x+5+x-3}{(x-3)(x+5)}=\frac{1}{3}$
$\Rightarrow \frac{2x+2}{x^{2}+5x-3x-15}=\frac{1}{3}$
$\Rightarrow \frac{2x+2}{x^{2}+2x-15}=\frac{1}{3}$
$\Rightarrow x^{2}+2x-15=3(2x+2)$
$\Rightarrow x^{2}+2x-15=6x+6$
$\Rightarrow x^{2}+2x-6x-15-6=0$
$\Rightarrow x^{2}-4x-21=0$
Here, $a=1$, $b=-4$, and $c=-21$
So, the discriminant, $b^{2}-4ac$
$=(-4)^{2}-4\times1\times(-21)$
$=16+84$
$=100>0$
Therefore:$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$=\frac{-(-4)\pm\sqrt{100}}{2\times1}$
$=\frac{4\pm10}{2}$
Either, $x=\frac{4+10}{2}$
$\Rightarrow x=\frac{14}{2}$
$\Rightarrow x=7$
or, $x=\frac{4-10}{2}$
$\Rightarrow x=\frac{-6}{2}$
$\Rightarrow x=-3$ [Not admissible as age cannot be negative]
So, the present age of Rehman is 7 years.
5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
Soln .
Let Shefali’s marks in Mathematics be $x$.
Then, her marks in English $=30-x$
According to the question:
$(x+2)\{(30-x)-3\}=210$
$\Rightarrow (x+2)(27-x)=210$
$\Rightarrow x(27-x)+2(27-x)=210$
$\Rightarrow 27x-x^{2}+54-2x=210$
$\Rightarrow -x^{2}+25x+54=210$
$\Rightarrow x^{2}-25x+210-54=0$
$\Rightarrow x^{2}-25x+156=0$
Here, $a=1$, $b=-25$, and $c=156$
So, the discriminant, $b^{2}-4ac$
$=(-25)^{2}-4\times1\times156$
$=625-624$
$=1>0$
Therefore:$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$=\frac{-(-25)\pm\sqrt{1}}{2\times1}$
$=\frac{25\pm1}{2}$
Either, $x=\frac{25+1}{2}$
$\Rightarrow x=\frac{26}{2}$
$\Rightarrow x=13$
or, $x=\frac{25-1}{2}$
$\Rightarrow x=\frac{24}{2}$
$\Rightarrow x=12$
If $x=12$, then marks in Mathematics $=13$ and marks in English $=30-13=17$
If $x=13$, then marks in Mathematics $=12$, and marks in English $=30-12=18$
So, Shefali’s marks in Mathematics and English are either 13 and 17 or 12 and 18.
6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Soln .
Let ABCD be the rectangular field, where BC be the shorter side, AB be the longer side and AC be the diagonal.
Let, $BC=x\text{ m}$
So, $AB=(x+30)\text{ m}$
$AC=(x+60)\text{ m}$
In right-angled $\triangle ABC$, we have
$AC^{2}=BC^{2}+AB^{2}$ [Pythagoras Theorem]
$\Rightarrow (x+60)^{2}=x^{2}+(x+30)^{2}$
$\Rightarrow x^{2}+120x+3600=x^{2}+x^{2}+60x+900$
$\Rightarrow x^{2}+120x+3600=2x^{2}+60x+900$
$\Rightarrow x^{2}+60x-120x+900-3600=0$
$\Rightarrow x^{2}-60x-2700=0$
Here, $a=1$, $b=-60$, and $c=-2700$
So, $b^{2}-4ac$
$=(-60)^{2}-4\times1\times(-2700)$
$=3600+10800$
$=14400>0$
Therefore:$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$=\frac{-(-60)\pm\sqrt{14400}}{2\times1}$
$=\frac{60\pm120}{2}$
Either, $x=\frac{60+120}{2}$
$\Rightarrow x=\frac{180}{2}$
$\Rightarrow x=90$
or, $x=\frac{60-120}{2}$
$\Rightarrow x=\frac{-60}{2}$
$\Rightarrow x=-30$ [Not admissible as length cannot be negative]
So, the shorter side is $90\text{ m}$ and the longer side is $(90+30)\text{ m}=120\text{ m}$.
7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Soln .
Let the larger number be $x$.
Then, the square of the smaller number $=8x$
Square of the larger number $=x^{2}$
According to the question:
$x^{2}-8x=180$
$\Rightarrow x^{2}-8x-180=0$
Here, $a=1$, $b=-8$, and $c=-180$
So, $b^{2}-4ac$
$=(-8)^{2}-4\times1\times(-180)$
$=64+720$
$=784>0$
Therefore:$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$=\frac{-(-8)\pm\sqrt{784}}{2\times1}$
$=\frac{8\pm28}{2}$
Either, $x=\frac{8+28}{2}$
$\Rightarrow x=\frac{36}{2}$
$\Rightarrow x=18$
or, $x=\frac{8-28}{2}$
$\Rightarrow x=\frac{-20}{2}$
$\Rightarrow x=-10$
Case I: When $x=18$,
Square of smaller number $=8\times18=144$
∴ Smaller number $=\sqrt{144}=\pm12$
Case II: When $x=-10$,
Square of smaller number $=8\times(-10)=-80$ [Not possible as the square of a real number cannot be negative]
So, the two numbers are 18, 12 or 18, -12.
8. If $\alpha$ and $\beta$ are the roots of the equation $3x^{2}+8x+2=0$, then the value of $\frac{1}{\alpha}+\frac{1}{\beta}$ is
(a) 4(b) -4(c) $\frac{2}{3}$(d) $\frac{-3}{8}$
Soln .
Given: $3x^{2}+8x+2=0$
Here, $a=3$, $b=8$, and $c=2$
We know that: Sum of roots,
$\alpha+\beta=-\frac{b}{a}=-\frac{8}{3}$
Product of roots, $\alpha\beta=\frac{c}{a}=\frac{2}{3}$
Now,$\frac{1}{\alpha}+\frac{1}{\beta}$
$=\frac{\alpha+\beta}{\alpha\beta}$
$=\frac{-\frac{8}{3}}{\frac{2}{3}}$
$=-\frac{8}{3}\times\frac{3}{2}$
$=-4$
Therefore, the correct option is (b) -4.