Question

v²-5v-24

o level,book d2​

Answer 1

Given,

v²-5v-24 = 0

Here we can use the formula:

$\boxed{\bold{\sf{v=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}}}$

Here,

• a = 1 {coefficient of v²}

• b = -5 {coefficient of v}

• c = -24 {coefficient of 1}

Applying the values to the equation:

$\sf{\implies v=\dfrac{-(-5)\pm\sqrt{(-5)^2-(4\times 1\times -24)}}{(2\times 1)}}$

$\sf{\implies v=\dfrac{5\pm\sqrt{25-(-96)}}{2}}$

$\sf{\implies v=\dfrac{5\pm\sqrt{25+96}}{2}}$

$\sf{\implies v=\dfrac{5\pm\sqrt{121}}{2}}$

$\sf{\implies v=\dfrac{5\pm11}{2}}$

$\sf{\implies v=\dfrac{5+11}{2} \ \ \ \ \ \& \ \ \ \ \ v=\dfrac{5-11}{2}}$

$\sf{\implies v=\dfrac{16}{2} \ \ \ \ \ \& \ \ \ \ \ v=\dfrac{-6}{2}}$

$\sf{\implies v=8 \ \ \ \ \ \& \ \ \ \ \ v=-3}$

$\boxed{\bold{\sf{\therefore \ The \ squares \ are \ 8 \ \& \ -3}}}$

Answer 2

Answer:

the answer is 8 and & -3

Step-by-step explanation:

the answer is 8 and & -3