Question

IF a - d, a, a+d are the Zeroes
2x² - 15x² + 37x-30 Find a and
b.​

Answer 1

(a - d), a, (a + d) are the zeroes of

2x³ - 15x² + 37x - 30

since, for any cubic polynomial ax³ + bx² + cx + d

hence,

(a - d) + a + (a + d) = $\frac{- b}{a} = \frac{15}{2}$

3a = $\frac{15}{2}$

a = $\frac{5}{2}$ ... (i)

now,

(a - d)(a) + (a)(a + d) + (a - d)(a + d) = $\frac{-d}{a} = 15$

a(a - d + a + d) + a² - d² = 15

2a² + a² - d² = 15

3a² - d² = 15

$3(\frac{25}{4}) - {d}^{2} = \frac{60}{4}$

${d}^{2} = \frac{75 + 60}{4} \\</p><p>{d}^{2} = \frac{135}{4} \\</p><p>d = ± \frac{3\sqrt{15}}{2}$

Answer 2

$\huge{ \green{ \mathfrak{ \underline{Question}}}}$

▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂

$If\: a - d,\: a, \:a+d\: are \:the\: zeroes\: of \\ 2x^3 - 15x^2 + 37x-30 \: , find\: a\: and\: b.$

▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂

$\huge{ \green{ \mathfrak{ \underline{Answer}}}}$

▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂

$\boxed{\red{\bold{Relationship \: of \: sum\:of\:zeroes:}}}$

$Sum \: of \: zeroes = \displaystyle{\frac{-Coefficient \: of \: x^2}{Coefficient \: of \:x^3}}$

$\implies (a-d)+a+(a+d) = \displaystyle{\frac{-(-15)}{2}}$

$\implies a-d + a + a+d = \displaystyle{\frac{15)}{2}}$

$\implies 3a = \displaystyle{\frac{-(-15)}{2}}$

$\implies a = \displaystyle{\frac{5)}{2}}$

$\implies a = 2.5$

$\boxed{\red{\bold{Relationship \: of \: product\:of\:zeroes:}}}$

$(a-d)a(a+d)=\displaystyle {\frac{-Constant\:term}{Coefficient \: of \:x^3}}$

$\implies (a^2-d^2)a = \displaystyle{\frac{-(-30)}{2}}$

$\implies [(2.5)^2 - d^2] × \displaystyle{\frac{5}{2}}= \displaystyle{\frac{30}{2}}$

$\implies 6.25 - d^2 = 15 \times \displaystyle{\frac{2}{5}}$

$\implies 6.25 - d^2 = 6$

$\implies d^2 = 6-6.25$

$\implies d = \pm \sqrt{0.25}$

$\implies d = \pm 0.5$

▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂