Question

Find the zeroes of the quadratic polynomial step by step..
ı) 3x^2-x-4
.ıı) t^2-15
^- SQUARE​

Answer 1

Given:-

๛Polynomials-p(x):-

➜3x²-x-4

➜t²-15

$\rule{200}{1}$

To Find:-

✪The Zeros of the polynomial-p(x)?

$\rule{200}{1}$

AnsWers:-

✪i)x=$\frac{4}{3}$

✪i)x=-1

★ii)x=±√15→√15 &-√15

$\rule{200}{1}$

➜i)3x²-x-4=0

↝S=-1→[Sum of Equation]

↝P=-12→[Product of Equation]

♦3-4=-1

♦3×-4=-12

↝3x²+3x-4x-4=0

↝3x(x+1)-4(x+1)=0

↝x+1=0

☞x=-1

↝3x-4=0

↝3x=4

☞x=$\frac{4}{3}$

$\rule{200}{1}$

➜ii)t²-15

↝t²-15=(t+√15)(t-√15)

↝t+√15=0

☞t=-√15

↝t-√15=0

☞t=√15

$\rule{200}{2}$

Answer 2

Answer:

$\underline{ \bf{\dag}\:\:\large{\textit{Question 1 :}}}$

$:\implies\sf f(x)=0\\\\\\:\implies\sf 3x^2-x-4=0\\\\\\:\implies\sf 3x^2-(4-3)x-4=0\\\\\\:\implies\sf 3x^2-4x+3x-4=0\\\\\\:\implies\sf x(3x-4)+1(3x-4)=0\\\\\\:\implies\sf (x+1)(3x-4)=0\\\\\\:\implies\underline{\boxed{\textsf{\textbf{x = - 1\quad or,\quad x = \dfrac{ \text4}{ \text3} }}}}$

$\rule{170}{2}$

$\underline{ \bf{\dag}\:\:\large{\textit{Question 2 :}}}$

$:\implies\sf f(x)=0\\\\\\:\implies\sf t^2-15=0\\\\\\:\implies\sf t^2=15\\\\\\:\implies\sf t=\sqrt{15}\\\\\\:\implies\sf t=\pm\:\sqrt{15}\\\\\\:\implies\underline{\boxed{\textsf{\textbf{t = \sqrt{\text{15}} \quad or,\quad t = - \sqrt{\text{15}} }}}}$