In each of the following, determine whether the given numbers are solutions of the given equation or not: (i) \[{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{3}\surd \mathbf{3x}\text{ }+\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0};\text{ }\mathbf{x}\text{ }=\text{ }\surd \mathbf{3},\text{ }-\mathbf{2}\surd \mathbf{3}\] (ii) \[{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\surd \mathbf{2x}\text{ }\text{ }\mathbf{4}\text{ }=\text{ }\mathbf{0};\text{ }\mathbf{x}\text{ }=\text{ }-\surd \mathbf{2},\text{ }\mathbf{2}\surd \mathbf{2}\]

In each of the following, determine whether the given numbers are solutions of the given equation or not: (i)

${{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{3}\surd \mathbf{3x}\text{ }+\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0};\text{ }\mathbf{x}\text{ }=\text{ }\surd \mathbf{3},\text{ }-\mathbf{2}\surd \mathbf{3}$

(ii)

${{\mathbf{x}}^{\mathbf{2}}}~\text{ }\surd \mathbf{2x}\text{ }\text{ }\mathbf{4}\text{ }=\text{ }\mathbf{0};\text{ }\mathbf{x}\text{ }=\text{ }-\surd \mathbf{2},\text{ }\mathbf{2}\surd \mathbf{2}$

In each of the following, determine whether the given numbers are solutions of the given equation or not: (i)

(ii)

${{\mathbf{x}}^{\mathbf{2}}}~\text{ }\surd \mathbf{2x}\text{ }\text{ }\mathbf{4}\text{ }=\text{ }\mathbf{0};\text{ }\mathbf{x}\text{ }=\text{ }-\surd \mathbf{2},\text{ }\mathbf{2}\surd \mathbf{2}$

(i)

Let us substitute the given values in the expression and check,

When,

$ (1) <img src="https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-037265abfbddb61039fcf638cce4ebba_l3.png" height="282" width="397" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*} & \begin{array}{*{35}{l}} x\text{ }=\text{ }\surd 3 \\ {{x}^{2}}~\text{ }3\surd 3x\text{ }+\text{ }6\text{ }=\text{ }0 \\ {{\left( \surd 3 \right)}^{2}}~\text{ }3\surd 3\left( \surd 3 \right)\text{ }+\text{ }6\text{ }=\text{ }0 \\ 3\text{ }\text{ }9\text{ }+\text{ }6\text{ }=\text{ }0 \\ -9\text{ }+\text{ }9\text{ }=0 \\ 0\text{ }=\text{ }0 \\ \end{array} \\ & \therefore \surd 3 \\ \end{align*}" title="Rendered by QuickLaTeX.com"/>$

is the solution of the equation.

When,

$ (2) <img src="https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-82554ee701d7140fc9ca9990ea76941d_l3.png" height="378" width="477" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*} & \begin{array}{*{35}{l}} x\text{ }=\text{ }-2\surd 3 \\ {{x}^{2}}~\text{ }3\surd 3x\text{ }+\text{ }6\text{ }=\text{ }0 \\ {{\left( -2\surd 3 \right)}^{2}}~\text{ }3\surd 3\left( -2\surd 3 \right)\text{ }+\text{ }6\text{ }=\text{ }0 \\ 4\left( 3 \right)\text{ }+18\text{ }+\text{ }6\text{ }=\text{ }0 \\ 12\text{ }+\text{ }18\text{ }+\text{ }6\text{ }=\text{ }0 \\ 36\text{ }=0 \\ \end{array} \\ & \\ & \\ & \therefore -2\surd 3 \\ \end{align*}" title="Rendered by QuickLaTeX.com"/>$

is not the solution of the equation.

(ii)

${{\mathbf{x}}^{\mathbf{2}}}~\text{ }\surd \mathbf{2x}\text{ }\text{ }\mathbf{4}\text{ }=\text{ }\mathbf{0};\text{ }\mathbf{x}\text{ }=\text{ }-\surd \mathbf{2},\text{ }\mathbf{2}\surd \mathbf{2}$

Let us substitute the given values in the expression and check,

When,

$ (3) <img src="https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ad19eb09fa98496db164b835d78a209e_l3.png" height="282" width="391" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*} & \begin{array}{*{35}{l}} x\text{ }=\text{ }-\surd 2 \\ {{x}^{2}}~\text{ }\surd 2x\text{ }\text{ }4\text{ }=\text{ }0 \\ {{\left( -\surd 2 \right)}^{2}}~\text{ }\surd 2\left( -\surd 2 \right)\text{ }\text{ }4\text{ }=\text{ }0 \\ 2\text{ }+\text{ }2\text{ }\text{ }4\text{ }=\text{ }0 \\ 4\text{ }\text{ }4\text{ }=\text{ }0 \\ 0\text{ }=\text{ }0 \\ \end{array} \\ & \therefore -\surd 2 \\ \end{align*}" title="Rendered by QuickLaTeX.com"/>$

is the solution of the equation.

When,

$ (4) <img src="https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1ff1191727ec361ef7fac703b5282513_l3.png" height="282" width="373" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*} & \begin{array}{*{35}{l}} x\text{ }=\text{ }2\surd 2 \\ {{x}^{2}}~\text{ }\surd 2x\text{ }\text{ }4\text{ }=\text{ }0 \\ {{\left( 2\surd 2 \right)}^{2}}~\text{ }\surd 2\left( 2\surd 2 \right)\text{ }\text{ }4\text{ }=\text{ }0 \\ 4\left( 2 \right)\text{ }\text{ }4\text{ }\text{ }4\text{ }=\text{ }0 \\ 4\text{ }\text{ }4\text{ }=\text{ }0 \\ 0\text{ }=\text{ }0 \\ \end{array} \\ & \therefore 2\surd 2 \\ \end{align*}" title="Rendered by QuickLaTeX.com"/>$

is the solution of the equation.

More Material

A …… lens is used to correct myopia and a ….. lens is used to correct hypermetropia.
A Concave, concave
B Convex, convex
C Convex, concave
D Concave, convex

Rate of reaction of any substance depends on
(A) active mass
(B) molecular weight
(C) atomic weight
(D) equivalent weight

Formula of oleum is
(A) $\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}$
(B) $\mathrm{H}_{2} \mathrm{SO}_{4}$
(C) $\mathrm{H}_{2} \mathrm{SO}_{3}$
(D) $\mathrm{H}_{2} \mathrm{SO}_{5}$

A line passes through the point (2, 1, -3) and is parallel to the vector Find the equations of the line in vector and Cartesian forms.

Two cells of emf $\varepsilon_{1}$ and $\varepsilon_{2}$ , internal resistance $r_{1}$ and $r_{2}$ , connected in parallel. The equivalent emf of the combination is- (A) $\frac{\varepsilon_{1} r_{1}+\varepsilon_{1} r_{1}}{r_{1}+r_{2}}$ (B) $\frac{\varepsilon_{1} r_{2}+\varepsilon_{2} r_{1}}{r_{1}+r_{2}}$ (C) $\sqrt{\varepsilon_{1} \times \varepsilon_{2}}$ (D) $\frac{\varepsilon_{1}+\varepsilon_{2}}{2}$

More Material

A …… lens is used to correct myopia and a ….. lens is used to correct hypermetropia.
A Concave, concave
B Convex, convex
C Convex, concave
D Concave, convex

Rate of reaction of any substance depends on
(A) active mass
(B) molecular weight
(C) atomic weight
(D) equivalent weight

Formula of oleum is
(A) $\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}$
(B) $\mathrm{H}_{2} \mathrm{SO}_{4}$
(C) $\mathrm{H}_{2} \mathrm{SO}_{3}$
(D) $\mathrm{H}_{2} \mathrm{SO}_{5}$

A line passes through the point (2, 1, -3) and is parallel to the vector Find the equations of the line in vector and Cartesian forms.

If the points $A(a, 0), B(0, b)$ and $C(1,1)$ are collinear, prove that $\frac{1}{a}+\frac{1}{b}=1$

If $A(-2,0), B(0,4)$ and $C(0, k)$ be three points such that area of a $A B C$ is 4 sq units, find the value of $k$ .

Find the value of $k$ for which the area of a ABC having vertices $A(2,-6), B(5,4)$ and $C(k, 4)$ is 35 sq units.

Find the value of $k$ for which the points $A(1,-1), B(2, k)$ and $C(4,5)$ are collinear.

Find the value of $k$ for which the points $P(5,5), Q(k, 1)$ and $R(11,7)$ are collinear.

Find the value of $k$ for which the points $A(3,-2), B(k, 2)$ and $C(8,8)$ are collinear.

Use determinants to show that the following points are collinear. $P(-2,5), Q(-6,-7)$ and $R(-5,-4)$

Sign up now and study all subjects for free

Class

Syllabus

Question Papers

Social Media

Apps