Question

Find the missing value to make the following expressions a perfect square trinomial.
1. a2 - 2a +___
2. p2 -- 5p +___
3. b² - 6b+___
4. x²-⅔x +___
plsss show me the solution
i mark you as a brainliest

Answer 1

Answer:

your answers are

Step-by-step explanation:

1. 1

2. 9

3. 6.25

4. 1/9

Answer 2

Given,

$\[\begin{array}{l}\left( 1 \right){a^2} - 2a + \_\_\\\left( 2 \right){p^2} - 5p + \_\_\\\left( 3 \right){b^2} - 6p + \_\_\\\left( 4 \right){x^2} - \frac{2}{3}x + \_\_\end{array}\]$

Solution,

Know that quadratic equation $\[a{x^2} + bx + c = 0\]$ has roots as a perfect square.

$\[\begin{array}{l}{b^2} - 4ac = 0\\ \Rightarrow {b^2} = 4ac\\ \Rightarrow c = \frac{{{b^2}}}{{4a}}\end{array}\]$

$\[\left( 1 \right){a^2} - 2a + \_\_\]$

$\[a = 1,b = - 2\]$

$\[\begin{array}{l}c = \frac{{{b^2}}}{{4a}}\\ \Rightarrow c = \frac{{{{\left( { - 2} \right)}^2}}}{{4 \times 1}}\\ \Rightarrow c = 1\end{array}\]$

Hence the expression is $\[{a^2} - 2a + 1\]$

$\[\left( 2 \right){p^2} - 5p + \_\_\]$

$\[\begin{array}{l}a = 1,b = - 5\\c = \frac{{{b^2}}}{{4a}}\\ \Rightarrow c = \frac{{{{\left( { - 5} \right)}^2}}}{{4 \times 1}}\\ \Rightarrow c = \frac{{25}}{4}\end{array}\]$

Hence the expression is $\[{p^2} - 5p + \frac{{25}}{4}\]$

$\[\left( 3 \right){b^2} - 6p + \_\_\]$

$\[\begin{array}{l}a = 1,b = - 6\\c = \frac{{{b^2}}}{{4a}}\\ \Rightarrow c = \frac{{{{\left( { - 6} \right)}^2}}}{{4 \times 1}}\\ \Rightarrow c = \frac{{36}}{4}\\ \Rightarrow c = 9\end{array}\]$

Hence the expression is $\[{b^2} - 6p + 9\]$

$\[\left( 4 \right){x^2} - \frac{2}{3}x + \_\_\]$

$\[\begin{array}{l}a = 1,b = - \frac{2}{3}\\c = \frac{{{b^2}}}{{4a}}\\ \Rightarrow c = \frac{{{{\left( { - \frac{2}{3}} \right)}^2}}}{{4 \times 1}}\\ \Rightarrow c = \frac{4}{{4 \times 9}}\\ \Rightarrow c = \frac{1}{9}\end{array}\]$

Hence the expression is $\[{x^2} - \frac{2}{3}x + \frac{1}{9}\]$.