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2x2 + x − 6
Completing the square,
{tex}\Rightarrow{/tex} x2 + {tex}\frac x2{/tex} − 3 = 0
{tex}\Rightarrow{/tex} x2 + {tex}\frac x2{/tex} + {tex}\frac 1{16}{/tex} − {tex}\frac 1{16}{/tex} − 3 = 0
{tex}\Rightarrow{/tex} (x + {tex}\frac 14{/tex})2 − {tex}\frac {49}{16}{/tex} = 0
{tex}\Rightarrow{/tex} (x + {tex}\frac 14{/tex} − {tex}\frac 74{/tex})(x + {tex}\frac 14{/tex} + {tex}\frac 74{/tex}) = 0
{tex}\Rightarrow{/tex} (x − {tex}\frac 32{/tex}) (x + 2) = 0
{tex}\Rightarrow{/tex} x = {tex}\frac 32{/tex}, −2
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If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.
Construction
Join the vertex B of {tex}\triangle{/tex}ABC to Q and the vertex C to P to form the lines BQ and CP and then drop a perpendicular QN to the side AB and also draw PM{tex}\perp{/tex}AC as shown in the given figure.
Proof
Now the area of {tex}\triangle{/tex}APQ = {tex}\frac {1}{2}{/tex} {tex}\times{/tex} AP {tex}\times{/tex} QN (Since, area of a triangle = {tex}\frac {1}{2}{/tex} {tex}\times{/tex} Base {tex}\times{/tex} Height)
Similarly, area of {tex}\triangle{/tex}PBQ = {tex}\frac {1}{2}{/tex} {tex}\times{/tex} PB {tex}\times{/tex} QN
area of {tex}\triangle{/tex}APQ = {tex}\frac {1}{2}{/tex} {tex}\times{/tex} AQ {tex}\times{/tex} PM
Also, area of {tex}\triangle{/tex}QCP = {tex}\frac {1}{2}{/tex} {tex}\times{/tex} QC {tex}\times{/tex} PM ...(i)
Now, if we find the ratio of the area of triangles {tex}\triangle{/tex}APQand {tex}\triangle{/tex}PBQ, we have
{tex}\frac{\text { area of } \Delta A P Q}{\text { area of } \Delta P B Q}{/tex} = {tex}\frac{\frac{1}{2} \times A P \times Q N}{\frac{1}{2} \times P B \times Q N}=\frac{A P}{P B}{/tex}
Similarly, {tex}\frac{\text { area of } \Delta A P Q}{\text { area of } \Delta Q C P}{/tex} = {tex}\frac{\frac{1}{2} \times A Q \times P M}{\frac{1}{2} \times Q C \times P M}=\frac{A Q}{Q C}{/tex} ...(ii)
According to the property of triangles, the triangles drawn between the same parallel lines and on the same base have equal areas.
Therefore, we can say that {tex}\triangle{/tex}PBQ and {tex}\triangle{/tex}QCP have the same area.
area of {tex}\triangle{/tex}PBQ = area of {tex}\triangle{/tex}QCP ...(iii)
Therefore, from the equations (i), (ii) and (iii) we can say that,
{tex}\frac{{AP}}{{PB}} = \frac{{AQ}}{{QC}}{/tex}
Also, {tex}\triangle{/tex}ABC and {tex}\triangle{/tex}APQ fulfil the conditions for similar triangles, as stated above. Thus, we can say that {tex}\triangle{/tex}ABC {tex} \sim {/tex} {tex}\triangle{/tex}APQ.
The MidPoint theorem is a special case of the basic proportionality theorem.
According to mid-point theorem, a line drawn joining the midpoints of the two sides of a triangle is parallel to the third side.
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Sia ? 4 years, 7 months ago
Logarithms are another way of thinking about exponents.
For example, we know that 2 raised to the 4th power equals 16. This is expressed by the exponential equation 24 = 16.
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