Question

solve the quadratic equation x^2-3√5x+10=0 AND -x^2+7x-10=0​

Answer 1

Step-by-step explanation:

Given quadratic equation:

x²-3√5x+10=0

Splitting the middle term, we

get

=> x²-2√5x-√5x+10=0

=> x²-2√5x-√5x+2×5=0

=> x²-2√5x+√5x +2×√5×√5=0

=> x(x-2√5)-√5(x-2√5)=0

=> (x-2√5)(x-√5)=0

=> x-2√5 =0 Or x-√5 = 0

=> x = 2√5 Or x = √5

Therefore,

Roots of given Quadratic equation are:

2√5 Or √5

.................................................

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Answer 2

Solution :-

(i) x² - 3√5 + 10

By using :-

$\star \sf{ x = \dfrac{ - b \pm \sqrt{b^2 - 4ac}}{2a} }$

here

• a = 1

• b = -3√5

• c = 10

$\small{ \implies \sf{ x = \dfrac{ -(-3\sqrt{5}) \pm \sqrt{(-3\sqrt{5})^2 - 4(1)(10)}}{2} } }$

$\implies \sf{ x = \dfrac{ 3\sqrt{5} \pm \sqrt{45 - 40}}{2} }$

$\implies \sf{ x = \dfrac{3\sqrt{5} \pm \sqrt{5}}{2} }$

So x = 2√5 or x = √5

(ii) -x² + 7x -10

By using :-

$\star \sf{ x = \dfrac{ - b \pm \sqrt{b^2 - 4ac}}{2a} }$

here

• a = -1

• b = 7

• c = -10

$\implies \sf{ x = \dfrac{ -(7) \pm \sqrt{(7)^2 - 4(-1)(-10)}}{-2} }$

$\implies \sf{ x = \dfrac{ -7 \pm \sqrt{49 - 40}}{-2} }$

$\implies \sf{ x = \dfrac{-7 \pm \sqrt{9}}{-2} }$

$\implies \sf{ x = \dfrac{-7 \pm 3}{-2} }$

So x = 2 or x = 5