Quadratic Equations
Category : 10th Class
Quadratic Equations
Quadratic Equations
Note: If \[\alpha \] is a root of the quadratic equation \[\mathbf{a}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+ bx + c = 0}\], then \[\alpha \] is called a zero of the polynomial \[\mathbf{a}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+ bx +c}\mathbf{.}\]
5. Nature of the Roots of a Quadratic Equation
Value of D |
Nature of Roots |
Real Roots |
\[D\text{ }>\text{ }0\] |
two distinct and real roots |
\[\alpha =\frac{-b\sqrt{D}}{2a}\,\,and\,\beta =\frac{-b-\sqrt{D}}{2a}\] |
\[D\text{ }=\text{ }0\] |
two equal real root |
\[each\,\,root\,\,\,=\,\,\,\frac{-b}{2a}\] |
\[D\text{ }<\text{ }0\] |
No real root |
None |
i.e., \[{{x}^{2}}-\text{ }\left( sum\text{ }of\text{ }roots \right)\text{ }x\text{ }+\text{ }\left( product\text{ }of\text{ }roots \right)\text{ }=\text{ }0\]
Snap Test
(a) \[k=\pm \,5\] (b) \[k=\pm \,8\]
(c) \[k=\pm \,3\] (d) \[k=\pm \,6\]
(e) None of these
Ans. (c)
Explanation: For equal roots discriminant D must be equal to 0.
In the given equation \[9{{x}^{2}}+\text{ }8kx\text{ }+\text{ }16\text{ }=\text{ }0\], we have:
\[D={{\left( 8k \right)}^{2}}-\text{ }(4\times 9\times 16)\]
Now, \[D\text{ }=\text{ }0\] \[\Rightarrow \] \[{{\left( 8k \right)}^{2}}-\text{ }(4\times ~9~\times 16)\text{ }=\text{ }0~~~[\,\therefore \,\,D\text{ }=\text{ }{{b}^{2}}-\text{ }4ac]\]
\[\Rightarrow \] \[64\text{ }\left\{ {{k}^{2}}-\text{ }9 \right\}=0~\,\,\Rightarrow ~\,\,{{k}^{2}}-\text{ }9\text{ }=\text{ }0~~\Rightarrow \,\,~\left( k\text{ }-\text{ }3 \right)\text{ }\left( k\text{ }+\text{ }3 \right)\text{ }=\text{ }0\]
\[\Rightarrow \] \[\left( k\text{ }-\text{ }3 \right)\text{ }=\text{ }0\text{ }or\text{ }\left( k\text{ }+\text{ }3 \right)\text{ }=\text{ }0~\,\,\Rightarrow \,\,k\text{ }=\text{ }3\text{ }or\text{ }k=-\text{ }3.\]
(a) \[a\text{ }=\text{ }3,\text{ }b\text{ =}\text{ }6\]
(b) \[a\text{ }=\text{ }3,\text{ }b\text{ }=+\text{ }6\]
(c) \[a\text{ }=\text{ }2,\text{ }b\text{ }=\text{ }-\text{ }5\]
(d) \[a\text{ }=\text{ }3,\text{ }b\text{ }=\text{ }-2\]
(e) None of these
Ans. (a)
Explanation: We have \[x\,\,=\,\,\frac{2}{3}\] for the equation \[a{{x}^{2}}+\text{ }7x\text{ }+\text{ }b\text{ }=\text{ }0\], therefore,
\[a{{\left( \frac{2}{3} \right)}^{2}}+7\left( \frac{2}{3} \right)+b=\,\,\Rightarrow \,\,4a+42+9b=0\,\,\Rightarrow \,\,4a+9b=-\,42\] …….. (i)
Since \[x=-\text{ }3\] is a root of the equation \[a{{x}^{2}}+\text{ }7x\text{ }+\text{ }b\text{ }=\text{ }0\], we have
\[a.\text{ }{{\left( -\text{ }3 \right)}^{2}}+\text{ }7\left( -\text{ }3 \right)\text{ }+\text{ }b\text{ }=\text{ }0\,\,\,\,\Rightarrow \,\,\,~9a\text{ }-\text{ }21\text{ }+\text{ }b\text{ }=\text{ }0~\,\,\,\Rightarrow \,\,\,9a\text{ }+\text{ }b\text{ }=\text{ }21\] …..... (ii)
Multiplying (ii) by 9 and subtracting (i) from it, we get:
\[77a\text{ }=\text{ }231\,\,\,\,\Rightarrow \,\,\,~a\text{ }=\text{ }3.\]
Substituting \[a\text{ }=\text{ }3\] in (i) we get: \[12+9b\text{ }=-\text{ }42~\,\,\,\Rightarrow \,\,\,9b=\,\,-54~\,\,\,\Rightarrow \,\,\,b=-\text{ }6\]
\[\frac{\mathbf{4}}{\mathbf{x}}\,\,\mathbf{-}\,\,\mathbf{3}\,\,\mathbf{=}\,\,\frac{\mathbf{5}}{\mathbf{2x+3}}\mathbf{,}\,\,\,\mathbf{x}\,\,\ne \,\,\mathbf{0,}\,\,\frac{\mathbf{-3}}{\mathbf{2}}\]
(a) \[{{x}^{2~}}-\text{ }x\text{ }+\text{ }1\text{ }=\text{ }0,\text{ }\left\{ -1,\text{ }1 \right\}\]
(b) \[{{x}^{2}}\text{ }x\text{ }+\text{ }2\text{ }=\text{ }0,\text{ }\left\{ -2,\text{ }1 \right\}\]
(c) \[{{x}^{2}}\text{ }x\text{ }+\text{ }4\text{ }=\text{ }0,\text{ }\left\{ -1,\text{ }2 \right\}\]
(d) \[x\text{ }+\text{ }x\text{ }+\text{ }2\text{ }=\text{ }0,\text{ }\left\{ -2,\text{ }1 \right\}\]
(e) None of these
Ans. (b)
Explanation: \[\frac{4}{x}\,\,-\,\,3\,\,\,=\,\,\,\,\frac{5}{2x+3},\,\,\,\,\Rightarrow \,\,\,\,\frac{4-3x}{x}\,\,=\,\,\frac{5}{2x+3}\]
\[\Rightarrow \] \[\left( 4\text{ }-\text{ }3x \right)\text{ }\left( 2x\text{ }+\text{ }3 \right)\text{ }=\text{ }5x~\,\,\,\,\Rightarrow \,\,\,-6{{x}^{2}}-\text{ }x\text{ }+\text{ }12\text{ }=\text{ }5x\]
\[\Rightarrow \] \[6{{x}^{2}}+\text{ }6x\text{ }-\text{ }12\text{ }=\text{ }0\,\,\,\,\Rightarrow \,\,\,~6\left( {{x}^{2}}+\text{ }x\text{ }-\text{ }2 \right)\text{ }=\text{ }0\]
\[\Rightarrow \] \[{{x}^{2}}+\text{ }x\text{ }-\text{ }2\text{ }=\text{ }0\], which is the required quadratic equation.
Now, \[{{x}^{2}}+\text{ }x\text{ }\text{ }2\text{ }=\text{ }0\,\,\,\Rightarrow \,\,~{{x}^{2}}+\text{ }2x\text{ }\text{ }x\text{ }\text{ }2\text{ }=\text{ }0\]
\[\Rightarrow \] \[x\text{ }(x+\text{ }2)\text{ }-1(x+\text{ }2)\text{ }=\text{ }0~\,\,\,\Rightarrow \,\,\,(x+2)(x-\text{ }1)\text{ }=\text{ }0\]
\[\Rightarrow \] \[\left( x\text{ }+\text{ }2 \right)=0\text{ }\,or\text{ }\left( x\text{ }-\text{ }1 \right)=0~\,\,\,\,\Rightarrow \,\,\,x=\,\,-2\text{ }or\text{ }x\text{ }=\text{ }1\]
(a) 45 (b) 36
(c) 12 (d) 24
(e) None of these
Ans. (d)
Explanation: Let the tens digit be a and the units digit be b.
Now, we have the required number \[=\text{ }10a\text{ }+\text{ }b\]
According to the question; \[10a\text{ }+\text{ }b\text{ }=\text{ }4\left( a\text{ }+\text{ }b \right)\,\,\,\,\Rightarrow \,\,\,~6a\text{ }=\text{ }3b\,\,\,\Rightarrow \,\,~b\text{ }=\text{ }2a\]
And \[10a\text{ }+\text{ }b+\text{ }18\text{ }=\text{ }10b\text{ }+\text{ }a\]
\[\Rightarrow \] \[9a\text{ }-\text{ }9b\text{ }+\text{ }18\text{ }=\text{ }0\]
\[\Rightarrow \] \[9a\text{ }-\text{ }18\text{ }a\text{ }+\text{ }18\text{ }=\text{ }0\]
\[\Rightarrow \] \[a\text{ }=\text{ }2\], thus \[b\text{ }=\text{ }4\]
(a) 84 (b) 36
(c) 34 (d) 93
(e) None of these
Ans. (a)
Explanation: Let tens digit be a and the units digit be b.
Now the required number \[=\text{ }10a\text{ }+\text{ }b\]
\[\Rightarrow \] \[10a\text{ }+\text{ }b\text{ }=\text{ }7\left( a\text{ }+\text{ }b \right)~~\Rightarrow \,\,\,3a\text{ }=\text{ }6b~\,\,\,\Rightarrow \,\,\,a\text{ }=\text{ }2b\]
And \[10a\text{ }+\text{ }b\text{ }=\text{ }3ab\text{ }\text{ }12\,\,\,\Rightarrow \,\,\,~20b\text{ }+\text{ }b\text{ }=\text{ }6{{b}^{2}}-\text{ }12\]
\[\Rightarrow \] \[6{{b}^{2}}-21b12=0\,\,\,\Rightarrow \,\,~3\left( 2{{b}^{2}}-7b-4 \right)=\text{ }0~\,\,\,\Rightarrow \,\,\,2{{b}^{2}}-7b4=0\]
\[\Rightarrow \] \[2{{b}^{2}}-\text{ }8b\text{ }+\text{ }b\text{ }\text{ }4\text{ }=\text{ }0\]
\[\Rightarrow \] \[2b\text{ }\left( b\text{ }-\text{ }4 \right)\text{ }+1\text{ }\left( b\text{ }-\text{ }4 \right)\text{ }=\text{ }0\,\,\,\Rightarrow \,\,~\left( b\text{ }-\text{ }4 \right)\text{ }\left( 2b\text{ }+\text{ }1 \right)\text{ }=\text{ }0\]
\[\Rightarrow \] \[b\text{ }=\text{ }4\] or \[b\,\,=\,\,\frac{-1}{2}\]
When \[b\text{ }=\text{ }4\], then \[a\text{ }=\text{ }8\]
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