Answer:

† Question :-

The quadratic equation 2x²– 8x + p = 0 has two distinct real roots. What is the highest possible integer value of p ?

\large \dag† Answer :-

\begin{gathered}\red\dashrightarrow\underline{\underline{\sf \green{Highest \: Possible \: Integer }} }\ \\ \green{\underline{\underline{\sf Value \: of \: p \: is \: 7 }}} \\\end{gathered}

⇢

Highest Possible Integer

Valueofpis7

\large \dag† Step by step EXPLANATION :-

We Know that If the roots are real & unequal of any quadratic equation ax² + bx + c then its discriminant is greater than zero i.e.

\begin{gathered}\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{ \blue{b^2-4AC > 0 }}}} \\ \end{gathered}

★

b

2

−4ac>0

Here We Have,

2x²– 8x + p = 0 has two distinct real roots

Therefore,

\begin{gathered} :\longmapsto \RM {( - 8)}^{2} - 4 \times 2 \times p > 0 \\ \end{gathered}

:⟼(−8)

2

−4×2×p>0

\begin{gathered}:\longmapsto \rm 64 - 8p > 0 \\ \end{gathered}

:⟼64−8p>0

\begin{gathered}:\longmapsto \rm 8p < 64 \\ \end{gathered}

:⟼8p<64

\begin{gathered}:\longmapsto \rm p < \cancel \frac{64}{8} \\ \end{gathered}

:⟼p<

8

64

\large \red{:\longmapsto} \green{ \underline{ \OVERLINE{ \pmb{ \boxed{ \rm p < 8}}}}}:⟼

p<8

p<8

So largest possible integer value of p LESS than 8 will be 7 .

Therefore,

\large\purple\leadsto\large\underline{\pink{\underline{\frak{\pmb{\text Answer \: is \: 7 }}}}}⇝

Answeris7

Answeris7

\large \dag† Additional Information :-

❒ Quadratic Polynomial with one Variable :

✪ The general form of the equation is ax² + bx + c = 0.

Note :

◆ If a = 0, then the equation becomes to a linear equation.

◆ If b = 0, then the roots of the equation becomes equal but opposite in sign.

◆ If c = 0, then one of the roots is zero. ]

❒ Nature Of Roots :

✪ b² - 4ac is the discriminant of the equation.

Then,

● If b² - 4ac = 0, then the roots are real & equal.

● If b² - 4ac > 0, then the roots are real & unequal.

● If b² - 4ac < 0, then the roots are imaginary