Question

Find the zeroes of the quadratic polynomial 6x^2+x-5

Answer 1

Step-by-step explanation:

Given -

• p(x) = 6x² + x - 5

To Find -

• Zeroes of the polynomial

Now,

→ 6x² + x - 5

By middle term splitt :-

→ 6x² + 6x - 5x - 5

→ 6x(x + 1) - 5(x + 1)

→ (6x - 5)(x + 1)

Zeroes are -

→ 6x - 5 = 0 and x + 1 = 0

→ x = 5/6 and x = -1

Verification :-

As we know that :-

• α + β = -b/a

→ -1 + 5/6 = -(1)/6

→ -6 + 5/6 = -1/6

→ -1/6 = -1/6

LHS = RHS

And

• αβ = c/a

→ -1 × 5/6 = -5/6

→ -5/6 = -5/6

LHS = RHS

Hence,

Verified...

Answer 2

$\star \; {\underline{\underline{\rm{\red{Given:}}}}}$

• p(x) = 6x² + x - 5

$\star \; {\underline{\underline{\rm{\red{To\;Find:}}}}}$

• Zeroes of the quadratic polynomial.

$\star \; {\underline{\underline{\rm{\red{Solution:}}}}}$

$: \implies$ 6x² + x - 5

$\small{\underline{\underline{\sf{\dag \; Splitting\;middle\;term:}}}}$

$: \implies$ 6x² + 6x - 5x - 5

$: \implies$ 6x(x + 1) -5(x + 1)

$: \implies$ (6x - 5)(x + 1)

$\small{\underline{\underline{\sf{\dag \; Therefore\;zeroes\;are:}}}}$

$: \leadsto$ (6x - 5) = 0

$: \leadsto \sf{x = \dfrac{5}{6}}$

$: \leadsto$ (x + 1) = 0

$: \leadsto$ x = -1

$\rule{200}{2}$

${\underline{\underline{\sf{\purple{\dag \; Verification:}}}}}$

We know that, if $\alpha$ and $\beta$ are zeroes of polynomial then:

$\small \; \; \; \; {\underline{\underline{\sf{\blue{\dag \; Sum\;of\;zeroes:}}}}}$

$\; \; \; \star \; \sf{ \alpha + \beta = \dfrac{-b}{a}}$

$: \implies \sf{ -1 + \dfrac{5}{6} = \dfrac{-1}{6}}$

$: \implies \sf{ \dfrac{-6 + 5}{6} = \dfrac{-1}{6}}$

$: \implies \sf{ \dfrac{-1}{6} = \dfrac{-1}{6}}$

$: \implies \sf{LHS = RHS}$

$\small \; \; \; \; {\underline{\underline{\sf{\blue{\dag \;Product\;of\;zeroes:}}}}}$

$\; \; \; \star \; \sf{ \alpha \beta = \dfrac{c}{a}}$

$: \implies \sf{ -1 \times \dfrac{5}{6} = \dfrac{-5}{6}}$

$: \implies \sf{ \dfrac{-5}{6} = \dfrac{-5}{6}}$

$: \implies \sf{LHS = RHS}$

${\underline{\underline{\sf{\purple{\dag \; Hence\; Verified!}}}}}$

$\rule{200}{2}$